Related
In Python remove() will remove the first occurrence of value in a list.
How to remove all occurrences of a value from a list?
This is what I have in mind:
>>> remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
[1, 3, 4, 3]
Functional approach:
Python 3.x
>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter((2).__ne__, x))
[1, 3, 3, 4]
or
>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter(lambda a: a != 2, x))
[1, 3, 3, 4]
or
>>> [i for i in x if i != 2]
Python 2.x
>>> x = [1,2,3,2,2,2,3,4]
>>> filter(lambda a: a != 2, x)
[1, 3, 3, 4]
You can use a list comprehension:
def remove_values_from_list(the_list, val):
return [value for value in the_list if value != val]
x = [1, 2, 3, 4, 2, 2, 3]
x = remove_values_from_list(x, 2)
print x
# [1, 3, 4, 3]
You can use slice assignment if the original list must be modified, while still using an efficient list comprehension (or generator expression).
>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> x[:] = (value for value in x if value != 2)
>>> x
[1, 3, 4, 3]
Repeating the solution of the first post in a more abstract way:
>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> while 2 in x: x.remove(2)
>>> x
[1, 3, 4, 3]
See the simple solution
>>> [i for i in x if i != 2]
This will return a list having all elements of x without 2
better solution with list comprehension
x = [ i for i in x if i!=2 ]
All of the answers above (apart from Martin Andersson's) create a new list without the desired items, rather than removing the items from the original list.
>>> import random, timeit
>>> a = list(range(5)) * 1000
>>> random.shuffle(a)
>>> b = a
>>> print(b is a)
True
>>> b = [x for x in b if x != 0]
>>> print(b is a)
False
>>> b.count(0)
0
>>> a.count(0)
1000
>>> b = a
>>> b = filter(lambda a: a != 2, x)
>>> print(b is a)
False
This can be important if you have other references to the list hanging around.
To modify the list in place, use a method like this
>>> def removeall_inplace(x, l):
... for _ in xrange(l.count(x)):
... l.remove(x)
...
>>> removeall_inplace(0, b)
>>> b is a
True
>>> a.count(0)
0
As far as speed is concerned, results on my laptop are (all on a 5000 entry list with 1000 entries removed)
List comprehension - ~400us
Filter - ~900us
.remove() loop - 50ms
So the .remove loop is about 100x slower........ Hmmm, maybe a different approach is needed. The fastest I've found is using the list comprehension, but then replace the contents of the original list.
>>> def removeall_replace(x, l):
.... t = [y for y in l if y != x]
.... del l[:]
.... l.extend(t)
removeall_replace() - 450us
Numpy approach and timings against a list/array with 1.000.000 elements:
Timings:
In [10]: a.shape
Out[10]: (1000000,)
In [13]: len(lst)
Out[13]: 1000000
In [18]: %timeit a[a != 2]
100 loops, best of 3: 2.94 ms per loop
In [19]: %timeit [x for x in lst if x != 2]
10 loops, best of 3: 79.7 ms per loop
Conclusion: numpy is 27 times faster (on my notebook) compared to list comprehension approach
PS if you want to convert your regular Python list lst to numpy array:
arr = np.array(lst)
Setup:
import numpy as np
a = np.random.randint(0, 1000, 10**6)
In [10]: a.shape
Out[10]: (1000000,)
In [12]: lst = a.tolist()
In [13]: len(lst)
Out[13]: 1000000
Check:
In [14]: a[a != 2].shape
Out[14]: (998949,)
In [15]: len([x for x in lst if x != 2])
Out[15]: 998949
At the cost of readability, I think this version is slightly faster as it doesn't force the while to reexamine the list, thus doing exactly the same work remove has to do anyway:
x = [1, 2, 3, 4, 2, 2, 3]
def remove_values_from_list(the_list, val):
for i in range(the_list.count(val)):
the_list.remove(val)
remove_values_from_list(x, 2)
print(x)
To remove all duplicate occurrences and leave one in the list:
test = [1, 1, 2, 3]
newlist = list(set(test))
print newlist
[1, 2, 3]
Here is the function I've used for Project Euler:
def removeOccurrences(e):
return list(set(e))
a = [1, 2, 2, 3, 1]
to_remove = 1
a = [i for i in a if i != to_remove]
print(a)
Perhaps not the most pythonic but still the easiest for me haha
for i in range(a.count(' ')):
a.remove(' ')
Much simpler I believe.
I believe this is probably faster than any other way if you don't care about the lists order, if you do take care about the final order store the indexes from the original and resort by that.
category_ids.sort()
ones_last_index = category_ids.count('1')
del category_ids[0:ones_last_index]
Let
>>> x = [1, 2, 3, 4, 2, 2, 3]
The simplest and efficient solution as already posted before is
>>> x[:] = [v for v in x if v != 2]
>>> x
[1, 3, 4, 3]
Another possibility which should use less memory but be slower is
>>> for i in range(len(x) - 1, -1, -1):
if x[i] == 2:
x.pop(i) # takes time ~ len(x) - i
>>> x
[1, 3, 4, 3]
Timing results for lists of length 1000 and 100000 with 10% matching entries: 0.16 vs 0.25 ms, and 23 vs 123 ms.
If your list contains only duplicates of only one element for example list_a=[0,0,0,0,0,0,1,3,4,6,7] the code below would be helpful:
list_a=[0,0,0,0,0,0,1,3,4,6,7]
def remove_element(element,the_list):
the_list=list(set(the_list))
the_list.remove(element)
return the_list
list_a=remove_element(element=0,the_list=list_a)
print(list_a)
or
a=list(set(i for i in list_a if i!=2))
a.remove(2)
The basic idea is that the sets do not allow duplicates, so first I have converted the list into set(which removes the duplicates), then used .remove() function to remove the first instance of the element(as now we have only one instance per item).
But if you have duplicates of multiple elements, the below methods would help:
List comprehension
list_a=[1, 2, 3, 4, 2, 2, 3]
remove_element=lambda element,the_list:[i for i in the_list if i!=element]
print(remove_element(element=2,the_list=list_a))
Filter
list_a=[1, 2, 3, 4, 2, 2, 3]
a=list(filter(lambda a: a != 2, list_a))
print(a)
While loop
list_a=[1, 2, 3, 4, 2, 2, 3]
def remove_element(element,the_list):
while element in the_list:the_list.remove(element)
return the_list
print(remove_element(2,list_a))
for loop (same as List comprehension)
list_a=[1, 2, 3, 4, 2, 2, 3]
a=[]
for i in list_a:
if i!=2:
a.append(i)
print(a)
Remove all occurrences of a value from a Python list
lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list():
for list in lists:
if(list!=7):
print(list)
remove_values_from_list()
Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11
Alternatively,
lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list(remove):
for list in lists:
if(list!=remove):
print(list)
remove_values_from_list(7)
Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11
I just did this for a list. I am just a beginner. A slightly more advanced programmer can surely write a function like this.
for i in range(len(spam)):
spam.remove('cat')
if 'cat' not in spam:
print('All instances of ' + 'cat ' + 'have been removed')
break
No one has posted an optimal answer for time and space complexity, so I thought I would give it a shot. Here is a solution that removes all occurrences of a specific value without creating a new array and at an efficient time complexity. The drawback is that the elements do not maintain order.
Time complexity: O(n)
Additional space complexity: O(1)
def main():
test_case([1, 2, 3, 4, 2, 2, 3], 2) # [1, 3, 3, 4]
test_case([3, 3, 3], 3) # []
test_case([1, 1, 1], 3) # [1, 1, 1]
def test_case(test_val, remove_val):
remove_element_in_place(test_val, remove_val)
print(test_val)
def remove_element_in_place(my_list, remove_value):
length_my_list = len(my_list)
swap_idx = length_my_list - 1
for idx in range(length_my_list - 1, -1, -1):
if my_list[idx] == remove_value:
my_list[idx], my_list[swap_idx] = my_list[swap_idx], my_list[idx]
swap_idx -= 1
for pop_idx in range(length_my_list - swap_idx - 1):
my_list.pop() # O(1) operation
if __name__ == '__main__':
main()
A lot of answers are really good. Here is a simple approach if you are a beginner in python in case you want to use the remove() method for sure.
rawlist = [8, 1, 8, 5, 8, 2, 8, 9, 8, 4]
ele_remove = 8
for el in rawlist:
if el == ele_remove:
rawlist.remove(ele_remove)
It may be slower for too large lists.
If you didn't have built-in filter or didn't want to use extra space and you need a linear solution...
def remove_all(A, v):
k = 0
n = len(A)
for i in range(n):
if A[i] != v:
A[k] = A[i]
k += 1
A = A[:k]
hello = ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']
#chech every item for a match
for item in range(len(hello)-1):
if hello[item] == ' ':
#if there is a match, rebuild the list with the list before the item + the list after the item
hello = hello[:item] + hello [item + 1:]
print hello
['h', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd']
We can also do in-place remove all using either del or pop:
import random
def remove_values_from_list(lst, target):
if type(lst) != list:
return lst
i = 0
while i < len(lst):
if lst[i] == target:
lst.pop(i) # length decreased by 1 already
else:
i += 1
return lst
remove_values_from_list(None, 2)
remove_values_from_list([], 2)
remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)], 2)
print(len(lst))
Now for the efficiency:
In [21]: %timeit -n1 -r1 x = random.randrange(0,10)
1 loop, best of 1: 43.5 us per loop
In [22]: %timeit -n1 -r1 lst = [random.randrange(0, 10) for x in range(1000000)]
g1 loop, best of 1: 660 ms per loop
In [23]: %timeit -n1 -r1 lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)]
...: , random.randrange(0,10))
1 loop, best of 1: 11.5 s per loop
In [27]: %timeit -n1 -r1 x = random.randrange(0,10); lst = [a for a in [random.randrange(0, 10) for x in
...: range(1000000)] if x != a]
1 loop, best of 1: 710 ms per loop
As we see that in-place version remove_values_from_list() does not require any extra memory, but it does take so much more time to run:
11 seconds for inplace remove values
710 milli seconds for list comprehensions, which allocates a new list in memory
You can convert your list to numpy.array and then use np.delete and pass the indices of the element and its all occurrences.
import numpy as np
my_list = [1, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7]
element_to_remove = 3
my_array = np.array(my_list)
indices = np.where(my_array == element_to_remove)
my_array = np.delete(my_array, indices)
my_list = my_array.tolist()
print(my_list)
#output
[1, 2, 4, 5, 6, 7, 4, 5, 6, 7]
About the speed!
import time
s_time = time.time()
print 'start'
a = range(100000000)
del a[:]
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 3.25
s_time = time.time()
print 'start'
a = range(100000000)
a = []
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 2.11
p=[2,3,4,4,4]
p.clear()
print(p)
[]
Only with Python 3
What's wrong with:
Motor=['1','2','2']
for i in Motor:
if i != '2':
print(i)
print(motor)
In Python remove() will remove the first occurrence of value in a list.
How to remove all occurrences of a value from a list?
This is what I have in mind:
>>> remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
[1, 3, 4, 3]
Functional approach:
Python 3.x
>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter((2).__ne__, x))
[1, 3, 3, 4]
or
>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter(lambda a: a != 2, x))
[1, 3, 3, 4]
or
>>> [i for i in x if i != 2]
Python 2.x
>>> x = [1,2,3,2,2,2,3,4]
>>> filter(lambda a: a != 2, x)
[1, 3, 3, 4]
You can use a list comprehension:
def remove_values_from_list(the_list, val):
return [value for value in the_list if value != val]
x = [1, 2, 3, 4, 2, 2, 3]
x = remove_values_from_list(x, 2)
print x
# [1, 3, 4, 3]
You can use slice assignment if the original list must be modified, while still using an efficient list comprehension (or generator expression).
>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> x[:] = (value for value in x if value != 2)
>>> x
[1, 3, 4, 3]
Repeating the solution of the first post in a more abstract way:
>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> while 2 in x: x.remove(2)
>>> x
[1, 3, 4, 3]
See the simple solution
>>> [i for i in x if i != 2]
This will return a list having all elements of x without 2
better solution with list comprehension
x = [ i for i in x if i!=2 ]
All of the answers above (apart from Martin Andersson's) create a new list without the desired items, rather than removing the items from the original list.
>>> import random, timeit
>>> a = list(range(5)) * 1000
>>> random.shuffle(a)
>>> b = a
>>> print(b is a)
True
>>> b = [x for x in b if x != 0]
>>> print(b is a)
False
>>> b.count(0)
0
>>> a.count(0)
1000
>>> b = a
>>> b = filter(lambda a: a != 2, x)
>>> print(b is a)
False
This can be important if you have other references to the list hanging around.
To modify the list in place, use a method like this
>>> def removeall_inplace(x, l):
... for _ in xrange(l.count(x)):
... l.remove(x)
...
>>> removeall_inplace(0, b)
>>> b is a
True
>>> a.count(0)
0
As far as speed is concerned, results on my laptop are (all on a 5000 entry list with 1000 entries removed)
List comprehension - ~400us
Filter - ~900us
.remove() loop - 50ms
So the .remove loop is about 100x slower........ Hmmm, maybe a different approach is needed. The fastest I've found is using the list comprehension, but then replace the contents of the original list.
>>> def removeall_replace(x, l):
.... t = [y for y in l if y != x]
.... del l[:]
.... l.extend(t)
removeall_replace() - 450us
Numpy approach and timings against a list/array with 1.000.000 elements:
Timings:
In [10]: a.shape
Out[10]: (1000000,)
In [13]: len(lst)
Out[13]: 1000000
In [18]: %timeit a[a != 2]
100 loops, best of 3: 2.94 ms per loop
In [19]: %timeit [x for x in lst if x != 2]
10 loops, best of 3: 79.7 ms per loop
Conclusion: numpy is 27 times faster (on my notebook) compared to list comprehension approach
PS if you want to convert your regular Python list lst to numpy array:
arr = np.array(lst)
Setup:
import numpy as np
a = np.random.randint(0, 1000, 10**6)
In [10]: a.shape
Out[10]: (1000000,)
In [12]: lst = a.tolist()
In [13]: len(lst)
Out[13]: 1000000
Check:
In [14]: a[a != 2].shape
Out[14]: (998949,)
In [15]: len([x for x in lst if x != 2])
Out[15]: 998949
At the cost of readability, I think this version is slightly faster as it doesn't force the while to reexamine the list, thus doing exactly the same work remove has to do anyway:
x = [1, 2, 3, 4, 2, 2, 3]
def remove_values_from_list(the_list, val):
for i in range(the_list.count(val)):
the_list.remove(val)
remove_values_from_list(x, 2)
print(x)
To remove all duplicate occurrences and leave one in the list:
test = [1, 1, 2, 3]
newlist = list(set(test))
print newlist
[1, 2, 3]
Here is the function I've used for Project Euler:
def removeOccurrences(e):
return list(set(e))
a = [1, 2, 2, 3, 1]
to_remove = 1
a = [i for i in a if i != to_remove]
print(a)
Perhaps not the most pythonic but still the easiest for me haha
for i in range(a.count(' ')):
a.remove(' ')
Much simpler I believe.
I believe this is probably faster than any other way if you don't care about the lists order, if you do take care about the final order store the indexes from the original and resort by that.
category_ids.sort()
ones_last_index = category_ids.count('1')
del category_ids[0:ones_last_index]
Let
>>> x = [1, 2, 3, 4, 2, 2, 3]
The simplest and efficient solution as already posted before is
>>> x[:] = [v for v in x if v != 2]
>>> x
[1, 3, 4, 3]
Another possibility which should use less memory but be slower is
>>> for i in range(len(x) - 1, -1, -1):
if x[i] == 2:
x.pop(i) # takes time ~ len(x) - i
>>> x
[1, 3, 4, 3]
Timing results for lists of length 1000 and 100000 with 10% matching entries: 0.16 vs 0.25 ms, and 23 vs 123 ms.
If your list contains only duplicates of only one element for example list_a=[0,0,0,0,0,0,1,3,4,6,7] the code below would be helpful:
list_a=[0,0,0,0,0,0,1,3,4,6,7]
def remove_element(element,the_list):
the_list=list(set(the_list))
the_list.remove(element)
return the_list
list_a=remove_element(element=0,the_list=list_a)
print(list_a)
or
a=list(set(i for i in list_a if i!=2))
a.remove(2)
The basic idea is that the sets do not allow duplicates, so first I have converted the list into set(which removes the duplicates), then used .remove() function to remove the first instance of the element(as now we have only one instance per item).
But if you have duplicates of multiple elements, the below methods would help:
List comprehension
list_a=[1, 2, 3, 4, 2, 2, 3]
remove_element=lambda element,the_list:[i for i in the_list if i!=element]
print(remove_element(element=2,the_list=list_a))
Filter
list_a=[1, 2, 3, 4, 2, 2, 3]
a=list(filter(lambda a: a != 2, list_a))
print(a)
While loop
list_a=[1, 2, 3, 4, 2, 2, 3]
def remove_element(element,the_list):
while element in the_list:the_list.remove(element)
return the_list
print(remove_element(2,list_a))
for loop (same as List comprehension)
list_a=[1, 2, 3, 4, 2, 2, 3]
a=[]
for i in list_a:
if i!=2:
a.append(i)
print(a)
Remove all occurrences of a value from a Python list
lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list():
for list in lists:
if(list!=7):
print(list)
remove_values_from_list()
Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11
Alternatively,
lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list(remove):
for list in lists:
if(list!=remove):
print(list)
remove_values_from_list(7)
Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11
I just did this for a list. I am just a beginner. A slightly more advanced programmer can surely write a function like this.
for i in range(len(spam)):
spam.remove('cat')
if 'cat' not in spam:
print('All instances of ' + 'cat ' + 'have been removed')
break
No one has posted an optimal answer for time and space complexity, so I thought I would give it a shot. Here is a solution that removes all occurrences of a specific value without creating a new array and at an efficient time complexity. The drawback is that the elements do not maintain order.
Time complexity: O(n)
Additional space complexity: O(1)
def main():
test_case([1, 2, 3, 4, 2, 2, 3], 2) # [1, 3, 3, 4]
test_case([3, 3, 3], 3) # []
test_case([1, 1, 1], 3) # [1, 1, 1]
def test_case(test_val, remove_val):
remove_element_in_place(test_val, remove_val)
print(test_val)
def remove_element_in_place(my_list, remove_value):
length_my_list = len(my_list)
swap_idx = length_my_list - 1
for idx in range(length_my_list - 1, -1, -1):
if my_list[idx] == remove_value:
my_list[idx], my_list[swap_idx] = my_list[swap_idx], my_list[idx]
swap_idx -= 1
for pop_idx in range(length_my_list - swap_idx - 1):
my_list.pop() # O(1) operation
if __name__ == '__main__':
main()
A lot of answers are really good. Here is a simple approach if you are a beginner in python in case you want to use the remove() method for sure.
rawlist = [8, 1, 8, 5, 8, 2, 8, 9, 8, 4]
ele_remove = 8
for el in rawlist:
if el == ele_remove:
rawlist.remove(ele_remove)
It may be slower for too large lists.
If you didn't have built-in filter or didn't want to use extra space and you need a linear solution...
def remove_all(A, v):
k = 0
n = len(A)
for i in range(n):
if A[i] != v:
A[k] = A[i]
k += 1
A = A[:k]
hello = ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']
#chech every item for a match
for item in range(len(hello)-1):
if hello[item] == ' ':
#if there is a match, rebuild the list with the list before the item + the list after the item
hello = hello[:item] + hello [item + 1:]
print hello
['h', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd']
We can also do in-place remove all using either del or pop:
import random
def remove_values_from_list(lst, target):
if type(lst) != list:
return lst
i = 0
while i < len(lst):
if lst[i] == target:
lst.pop(i) # length decreased by 1 already
else:
i += 1
return lst
remove_values_from_list(None, 2)
remove_values_from_list([], 2)
remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)], 2)
print(len(lst))
Now for the efficiency:
In [21]: %timeit -n1 -r1 x = random.randrange(0,10)
1 loop, best of 1: 43.5 us per loop
In [22]: %timeit -n1 -r1 lst = [random.randrange(0, 10) for x in range(1000000)]
g1 loop, best of 1: 660 ms per loop
In [23]: %timeit -n1 -r1 lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)]
...: , random.randrange(0,10))
1 loop, best of 1: 11.5 s per loop
In [27]: %timeit -n1 -r1 x = random.randrange(0,10); lst = [a for a in [random.randrange(0, 10) for x in
...: range(1000000)] if x != a]
1 loop, best of 1: 710 ms per loop
As we see that in-place version remove_values_from_list() does not require any extra memory, but it does take so much more time to run:
11 seconds for inplace remove values
710 milli seconds for list comprehensions, which allocates a new list in memory
You can convert your list to numpy.array and then use np.delete and pass the indices of the element and its all occurrences.
import numpy as np
my_list = [1, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7]
element_to_remove = 3
my_array = np.array(my_list)
indices = np.where(my_array == element_to_remove)
my_array = np.delete(my_array, indices)
my_list = my_array.tolist()
print(my_list)
#output
[1, 2, 4, 5, 6, 7, 4, 5, 6, 7]
About the speed!
import time
s_time = time.time()
print 'start'
a = range(100000000)
del a[:]
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 3.25
s_time = time.time()
print 'start'
a = range(100000000)
a = []
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 2.11
p=[2,3,4,4,4]
p.clear()
print(p)
[]
Only with Python 3
What's wrong with:
Motor=['1','2','2']
for i in Motor:
if i != '2':
print(i)
print(motor)
I want to get the rank of each element, so I use argsort in numpy:
np.argsort(np.array((1,1,1,2,2,3,3,3,3)))
array([0, 1, 2, 3, 4, 5, 6, 7, 8])
it give the same element the different rank, can I get the same rank like:
array([0, 0, 0, 3, 3, 5, 5, 5, 5])
If you don't mind a dependency on scipy, you can use scipy.stats.rankdata, with method='min':
In [14]: a
Out[14]: array([1, 1, 1, 2, 2, 3, 3, 3, 3])
In [15]: from scipy.stats import rankdata
In [16]: rankdata(a, method='min')
Out[16]: array([1, 1, 1, 4, 4, 6, 6, 6, 6])
Note that rankdata starts the ranks at 1. To start at 0, subtract 1 from the result:
In [17]: rankdata(a, method='min') - 1
Out[17]: array([0, 0, 0, 3, 3, 5, 5, 5, 5])
If you don't want the scipy dependency, you can use numpy.unique to compute the ranking. Here's a function that computes the same result as rankdata(x, method='min') - 1:
import numpy as np
def rankmin(x):
u, inv, counts = np.unique(x, return_inverse=True, return_counts=True)
csum = np.zeros_like(counts)
csum[1:] = counts[:-1].cumsum()
return csum[inv]
For example,
In [137]: x = np.array([60, 10, 0, 30, 20, 40, 50])
In [138]: rankdata(x, method='min') - 1
Out[138]: array([6, 1, 0, 3, 2, 4, 5])
In [139]: rankmin(x)
Out[139]: array([6, 1, 0, 3, 2, 4, 5])
In [140]: a = np.array([1,1,1,2,2,3,3,3,3])
In [141]: rankdata(a, method='min') - 1
Out[141]: array([0, 0, 0, 3, 3, 5, 5, 5, 5])
In [142]: rankmin(a)
Out[142]: array([0, 0, 0, 3, 3, 5, 5, 5, 5])
By the way, a single call to argsort() does not give ranks. You can find an assortment of approaches to ranking in the question Rank items in an array using Python/NumPy, including how to do it using argsort().
Alternatively, pandas series has a rank method which does what you need with the min method:
import pandas as pd
pd.Series((1,1,1,2,2,3,3,3,3)).rank(method="min")
# 0 1
# 1 1
# 2 1
# 3 4
# 4 4
# 5 6
# 6 6
# 7 6
# 8 6
# dtype: float64
With focus on performance, here's an approach -
def rank_repeat_based(arr):
idx = np.concatenate(([0],np.flatnonzero(np.diff(arr))+1,[arr.size]))
return np.repeat(idx[:-1],np.diff(idx))
For a generic case with the elements in input array not already sorted, we would need to use argsort() to keep track of the positions. So, we would have a modified version, like so -
def rank_repeat_based_generic(arr):
sidx = np.argsort(arr,kind='mergesort')
idx = np.concatenate(([0],np.flatnonzero(np.diff(arr[sidx]))+1,[arr.size]))
return np.repeat(idx[:-1],np.diff(idx))[sidx.argsort()]
Runtime test
Testing out all the approaches listed thus far to solve the problem on a large dataset.
Sorted array case :
In [96]: arr = np.sort(np.random.randint(1,100,(10000)))
In [97]: %timeit rankdata(arr, method='min') - 1
1000 loops, best of 3: 635 µs per loop
In [98]: %timeit rankmin(arr)
1000 loops, best of 3: 495 µs per loop
In [99]: %timeit (pd.Series(arr).rank(method="min")-1).values
1000 loops, best of 3: 826 µs per loop
In [100]: %timeit rank_repeat_based(arr)
10000 loops, best of 3: 200 µs per loop
Unsorted case :
In [106]: arr = np.random.randint(1,100,(10000))
In [107]: %timeit rankdata(arr, method='min') - 1
1000 loops, best of 3: 963 µs per loop
In [108]: %timeit rankmin(arr)
1000 loops, best of 3: 869 µs per loop
In [109]: %timeit (pd.Series(arr).rank(method="min")-1).values
1000 loops, best of 3: 1.17 ms per loop
In [110]: %timeit rank_repeat_based_generic(arr)
1000 loops, best of 3: 1.76 ms per loop
I've written a function for the same purpose. It uses pure python and numpy only. Please have a look. I put comments as well.
def my_argsort(array):
# this type conversion let us work with python lists and pandas series
array = np.array(array)
# create mapping for unique values
# it's a dictionary where keys are values from the array and
# values are desired indices
unique_values = list(set(array))
mapping = dict(zip(unique_values, np.argsort(unique_values)))
# apply mapping to our array
# np.vectorize works similar map(), and can work with dictionaries
array = np.vectorize(mapping.get)(array)
return array
Hope that helps.
Complex solutions are unnecessary for this problem.
> ary = np.sort([1, 1, 1, 2, 2, 3, 3, 3, 3]) # or anything; must be sorted.
> a = np.diff().cumsum(); a
array([0, 0, 1, 1, 2, 2, 2, 2])
> b = np.r_[0, a]; b # ties get first open rank
array([0, 0, 0, 1, 1, 2, 2, 2, 2])
> c = np.flatnonzero(ary[1:] != ary[:-1])
> np.r_[0, 1 + c][b] # ties get last open rank
array([0, 0, 0, 3, 3, 5, 5, 5])
I have this:
>>> a = [1, 2, 4]
>>> print a
[1, 2, 4]
>>> print a.insert(2, 3)
None
>>> print a
[1, 2, 3, 4]
>>> b = a.insert(3, 6)
>>> print b
None
>>> print a
[1, 2, 3, 6, 4]
Is there a way I can get the updated list as the result, instead of updating the original list in place?
l.insert(index, obj) doesn't actually return anything. It just updates the list.
As ATO said, you can do b = a[:index] + [obj] + a[index:].
However, another way is:
a = [1, 2, 4]
b = a[:]
b.insert(2, 3)
Most performance efficient approach
You may also insert the element using the slice indexing in the list. For example:
>>> a = [1, 2, 4]
>>> insert_at = 2 # Index at which you want to insert item
>>> b = a[:] # Created copy of list "a" as "b".
# Skip this step if you are ok with modifying the original list
>>> b[insert_at:insert_at] = [3] # Insert "3" within "b"
>>> b
[1, 2, 3, 4]
For inserting multiple elements together at a given index, all you need to do is to use a list of multiple elements that you want to insert. For example:
>>> a = [1, 2, 4]
>>> insert_at = 2 # Index starting from which multiple elements will be inserted
# List of elements that you want to insert together at "index_at" (above) position
>>> insert_elements = [3, 5, 6]
>>> a[insert_at:insert_at] = insert_elements
>>> a # [3, 5, 6] are inserted together in `a` starting at index "2"
[1, 2, 3, 5, 6, 4]
To know more about slice indexing, you can refer: Understanding slice notation.
Note: In Python 3.x, difference of performance between slice indexing and list.index(...) is significantly reduced and both are almost equivalent. However, in Python 2.x, this difference is quite noticeable. I have shared performance comparisons later in this answer.
Alternative using list comprehension (but very slow in terms of performance):
As an alternative, it can be achieved using list comprehension with enumerate too. (But please don't do it this way. It is just for illustration):
>>> a = [1, 2, 4]
>>> insert_at = 2
>>> b = [y for i, x in enumerate(a) for y in ((3, x) if i == insert_at else (x, ))]
>>> b
[1, 2, 3, 4]
Performance comparison of all solutions
Here's the timeit comparison of all the answers with list of 1000 elements on Python 3.9.1 and Python 2.7.16. Answers are listed in the order of performance for both the Python versions.
Python 3.9.1
My answer using sliced insertion - Fastest ( 2.25 µsec per loop)
python3 -m timeit -s "a = list(range(1000))" "b = a[:]; b[500:500] = [3]"
100000 loops, best of 5: 2.25 µsec per loop
Rushy Panchal's answer with most votes using list.insert(...)- Second (2.33 µsec per loop)
python3 -m timeit -s "a = list(range(1000))" "b = a[:]; b.insert(500, 3)"
100000 loops, best of 5: 2.33 µsec per loop
ATOzTOA's accepted answer based on merge of sliced lists - Third (5.01 µsec per loop)
python3 -m timeit -s "a = list(range(1000))" "b = a[:500] + [3] + a[500:]"
50000 loops, best of 5: 5.01 µsec per loop
My answer with List Comprehension and enumerate - Fourth (very slow with 135 µsec per loop)
python3 -m timeit -s "a = list(range(1000))" "[y for i, x in enumerate(a) for y in ((3, x) if i == 500 else (x, )) ]"
2000 loops, best of 5: 135 µsec per loop
Python 2.7.16
My answer using sliced insertion - Fastest (2.09 µsec per loop)
python -m timeit -s "a = list(range(1000))" "b = a[:]; b[500:500] = [3]"
100000 loops, best of 3: 2.09 µsec per loop
Rushy Panchal's answer with most votes using list.insert(...)- Second (2.36 µsec per loop)
python -m timeit -s "a = list(range(1000))" "b = a[:]; b.insert(500, 3)"
100000 loops, best of 3: 2.36 µsec per loop
ATOzTOA's accepted answer based on merge of sliced lists - Third (4.44 µsec per loop)
python -m timeit -s "a = list(range(1000))" "b = a[:500] + [3] + a[500:]"
100000 loops, best of 3: 4.44 µsec per loop
My answer with List Comprehension and enumerate - Fourth (very slow with 103 µsec per loop)
python -m timeit -s "a = list(range(1000))" "[y for i, x in enumerate(a) for y in ((3, x) if i == 500 else (x, )) ]"
10000 loops, best of 3: 103 µsec per loop
The shortest I got: b = a[:2] + [3] + a[2:]
>>>
>>> a = [1, 2, 4]
>>> print a
[1, 2, 4]
>>> b = a[:2] + [3] + a[2:]
>>> print a
[1, 2, 4]
>>> print b
[1, 2, 3, 4]
The cleanest approach is to copy the list and then insert the object into the copy. On Python 3 this can be done via list.copy:
new = old.copy()
new.insert(index, value)
On Python 2 copying the list can be achieved via new = old[:] (this also works on Python 3).
In terms of performance there is no difference to other proposed methods:
$ python --version
Python 3.8.1
$ python -m timeit -s "a = list(range(1000))" "b = a.copy(); b.insert(500, 3)"
100000 loops, best of 5: 2.84 µsec per loop
$ python -m timeit -s "a = list(range(1000))" "b = a.copy(); b[500:500] = (3,)"
100000 loops, best of 5: 2.76 µsec per loop
Here is the way to add a single item, a single item in specific index concatenate list with another list
>>> expences = [2200, 2350, 2600, 2130, 2190]
>>> expences.append(1980)
>>> expences
[2200, 2350, 2600, 2130, 2190, 1980]
>>> expences.insert(1, 1200)
>>> expences
[2200, 1200, 2350, 2600, 2130, 2190, 1980]
>>> newElm = [2550, 2123, 2430]
>>> expences.extend(newElm)
>>> expences
[2200, 1200, 2350, 2600, 2130, 2190, 1980, 2550, 2123, 2430]
>>>
Use the Python list insert() method. Usage:
#Syntax
The syntax for the insert() method −
list.insert(index, obj)
#Parameters
index − This is the Index where the object obj need to be inserted.
obj − This is the Object to be inserted into the given list.
#Return Value
This method does not return any value, but it inserts the given element at the given index.
Example:
a = [1,2,4,5]
a.insert(2,3)
print(a)
Returns [1, 2, 3, 4, 5]
In Python remove() will remove the first occurrence of value in a list.
How to remove all occurrences of a value from a list?
This is what I have in mind:
>>> remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
[1, 3, 4, 3]
Functional approach:
Python 3.x
>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter((2).__ne__, x))
[1, 3, 3, 4]
or
>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter(lambda a: a != 2, x))
[1, 3, 3, 4]
or
>>> [i for i in x if i != 2]
Python 2.x
>>> x = [1,2,3,2,2,2,3,4]
>>> filter(lambda a: a != 2, x)
[1, 3, 3, 4]
You can use a list comprehension:
def remove_values_from_list(the_list, val):
return [value for value in the_list if value != val]
x = [1, 2, 3, 4, 2, 2, 3]
x = remove_values_from_list(x, 2)
print x
# [1, 3, 4, 3]
You can use slice assignment if the original list must be modified, while still using an efficient list comprehension (or generator expression).
>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> x[:] = (value for value in x if value != 2)
>>> x
[1, 3, 4, 3]
Repeating the solution of the first post in a more abstract way:
>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> while 2 in x: x.remove(2)
>>> x
[1, 3, 4, 3]
See the simple solution
>>> [i for i in x if i != 2]
This will return a list having all elements of x without 2
better solution with list comprehension
x = [ i for i in x if i!=2 ]
All of the answers above (apart from Martin Andersson's) create a new list without the desired items, rather than removing the items from the original list.
>>> import random, timeit
>>> a = list(range(5)) * 1000
>>> random.shuffle(a)
>>> b = a
>>> print(b is a)
True
>>> b = [x for x in b if x != 0]
>>> print(b is a)
False
>>> b.count(0)
0
>>> a.count(0)
1000
>>> b = a
>>> b = filter(lambda a: a != 2, x)
>>> print(b is a)
False
This can be important if you have other references to the list hanging around.
To modify the list in place, use a method like this
>>> def removeall_inplace(x, l):
... for _ in xrange(l.count(x)):
... l.remove(x)
...
>>> removeall_inplace(0, b)
>>> b is a
True
>>> a.count(0)
0
As far as speed is concerned, results on my laptop are (all on a 5000 entry list with 1000 entries removed)
List comprehension - ~400us
Filter - ~900us
.remove() loop - 50ms
So the .remove loop is about 100x slower........ Hmmm, maybe a different approach is needed. The fastest I've found is using the list comprehension, but then replace the contents of the original list.
>>> def removeall_replace(x, l):
.... t = [y for y in l if y != x]
.... del l[:]
.... l.extend(t)
removeall_replace() - 450us
Numpy approach and timings against a list/array with 1.000.000 elements:
Timings:
In [10]: a.shape
Out[10]: (1000000,)
In [13]: len(lst)
Out[13]: 1000000
In [18]: %timeit a[a != 2]
100 loops, best of 3: 2.94 ms per loop
In [19]: %timeit [x for x in lst if x != 2]
10 loops, best of 3: 79.7 ms per loop
Conclusion: numpy is 27 times faster (on my notebook) compared to list comprehension approach
PS if you want to convert your regular Python list lst to numpy array:
arr = np.array(lst)
Setup:
import numpy as np
a = np.random.randint(0, 1000, 10**6)
In [10]: a.shape
Out[10]: (1000000,)
In [12]: lst = a.tolist()
In [13]: len(lst)
Out[13]: 1000000
Check:
In [14]: a[a != 2].shape
Out[14]: (998949,)
In [15]: len([x for x in lst if x != 2])
Out[15]: 998949
At the cost of readability, I think this version is slightly faster as it doesn't force the while to reexamine the list, thus doing exactly the same work remove has to do anyway:
x = [1, 2, 3, 4, 2, 2, 3]
def remove_values_from_list(the_list, val):
for i in range(the_list.count(val)):
the_list.remove(val)
remove_values_from_list(x, 2)
print(x)
To remove all duplicate occurrences and leave one in the list:
test = [1, 1, 2, 3]
newlist = list(set(test))
print newlist
[1, 2, 3]
Here is the function I've used for Project Euler:
def removeOccurrences(e):
return list(set(e))
a = [1, 2, 2, 3, 1]
to_remove = 1
a = [i for i in a if i != to_remove]
print(a)
Perhaps not the most pythonic but still the easiest for me haha
for i in range(a.count(' ')):
a.remove(' ')
Much simpler I believe.
I believe this is probably faster than any other way if you don't care about the lists order, if you do take care about the final order store the indexes from the original and resort by that.
category_ids.sort()
ones_last_index = category_ids.count('1')
del category_ids[0:ones_last_index]
Let
>>> x = [1, 2, 3, 4, 2, 2, 3]
The simplest and efficient solution as already posted before is
>>> x[:] = [v for v in x if v != 2]
>>> x
[1, 3, 4, 3]
Another possibility which should use less memory but be slower is
>>> for i in range(len(x) - 1, -1, -1):
if x[i] == 2:
x.pop(i) # takes time ~ len(x) - i
>>> x
[1, 3, 4, 3]
Timing results for lists of length 1000 and 100000 with 10% matching entries: 0.16 vs 0.25 ms, and 23 vs 123 ms.
If your list contains only duplicates of only one element for example list_a=[0,0,0,0,0,0,1,3,4,6,7] the code below would be helpful:
list_a=[0,0,0,0,0,0,1,3,4,6,7]
def remove_element(element,the_list):
the_list=list(set(the_list))
the_list.remove(element)
return the_list
list_a=remove_element(element=0,the_list=list_a)
print(list_a)
or
a=list(set(i for i in list_a if i!=2))
a.remove(2)
The basic idea is that the sets do not allow duplicates, so first I have converted the list into set(which removes the duplicates), then used .remove() function to remove the first instance of the element(as now we have only one instance per item).
But if you have duplicates of multiple elements, the below methods would help:
List comprehension
list_a=[1, 2, 3, 4, 2, 2, 3]
remove_element=lambda element,the_list:[i for i in the_list if i!=element]
print(remove_element(element=2,the_list=list_a))
Filter
list_a=[1, 2, 3, 4, 2, 2, 3]
a=list(filter(lambda a: a != 2, list_a))
print(a)
While loop
list_a=[1, 2, 3, 4, 2, 2, 3]
def remove_element(element,the_list):
while element in the_list:the_list.remove(element)
return the_list
print(remove_element(2,list_a))
for loop (same as List comprehension)
list_a=[1, 2, 3, 4, 2, 2, 3]
a=[]
for i in list_a:
if i!=2:
a.append(i)
print(a)
Remove all occurrences of a value from a Python list
lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list():
for list in lists:
if(list!=7):
print(list)
remove_values_from_list()
Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11
Alternatively,
lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list(remove):
for list in lists:
if(list!=remove):
print(list)
remove_values_from_list(7)
Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11
I just did this for a list. I am just a beginner. A slightly more advanced programmer can surely write a function like this.
for i in range(len(spam)):
spam.remove('cat')
if 'cat' not in spam:
print('All instances of ' + 'cat ' + 'have been removed')
break
No one has posted an optimal answer for time and space complexity, so I thought I would give it a shot. Here is a solution that removes all occurrences of a specific value without creating a new array and at an efficient time complexity. The drawback is that the elements do not maintain order.
Time complexity: O(n)
Additional space complexity: O(1)
def main():
test_case([1, 2, 3, 4, 2, 2, 3], 2) # [1, 3, 3, 4]
test_case([3, 3, 3], 3) # []
test_case([1, 1, 1], 3) # [1, 1, 1]
def test_case(test_val, remove_val):
remove_element_in_place(test_val, remove_val)
print(test_val)
def remove_element_in_place(my_list, remove_value):
length_my_list = len(my_list)
swap_idx = length_my_list - 1
for idx in range(length_my_list - 1, -1, -1):
if my_list[idx] == remove_value:
my_list[idx], my_list[swap_idx] = my_list[swap_idx], my_list[idx]
swap_idx -= 1
for pop_idx in range(length_my_list - swap_idx - 1):
my_list.pop() # O(1) operation
if __name__ == '__main__':
main()
A lot of answers are really good. Here is a simple approach if you are a beginner in python in case you want to use the remove() method for sure.
rawlist = [8, 1, 8, 5, 8, 2, 8, 9, 8, 4]
ele_remove = 8
for el in rawlist:
if el == ele_remove:
rawlist.remove(ele_remove)
It may be slower for too large lists.
If you didn't have built-in filter or didn't want to use extra space and you need a linear solution...
def remove_all(A, v):
k = 0
n = len(A)
for i in range(n):
if A[i] != v:
A[k] = A[i]
k += 1
A = A[:k]
hello = ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']
#chech every item for a match
for item in range(len(hello)-1):
if hello[item] == ' ':
#if there is a match, rebuild the list with the list before the item + the list after the item
hello = hello[:item] + hello [item + 1:]
print hello
['h', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd']
We can also do in-place remove all using either del or pop:
import random
def remove_values_from_list(lst, target):
if type(lst) != list:
return lst
i = 0
while i < len(lst):
if lst[i] == target:
lst.pop(i) # length decreased by 1 already
else:
i += 1
return lst
remove_values_from_list(None, 2)
remove_values_from_list([], 2)
remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)], 2)
print(len(lst))
Now for the efficiency:
In [21]: %timeit -n1 -r1 x = random.randrange(0,10)
1 loop, best of 1: 43.5 us per loop
In [22]: %timeit -n1 -r1 lst = [random.randrange(0, 10) for x in range(1000000)]
g1 loop, best of 1: 660 ms per loop
In [23]: %timeit -n1 -r1 lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)]
...: , random.randrange(0,10))
1 loop, best of 1: 11.5 s per loop
In [27]: %timeit -n1 -r1 x = random.randrange(0,10); lst = [a for a in [random.randrange(0, 10) for x in
...: range(1000000)] if x != a]
1 loop, best of 1: 710 ms per loop
As we see that in-place version remove_values_from_list() does not require any extra memory, but it does take so much more time to run:
11 seconds for inplace remove values
710 milli seconds for list comprehensions, which allocates a new list in memory
You can convert your list to numpy.array and then use np.delete and pass the indices of the element and its all occurrences.
import numpy as np
my_list = [1, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7]
element_to_remove = 3
my_array = np.array(my_list)
indices = np.where(my_array == element_to_remove)
my_array = np.delete(my_array, indices)
my_list = my_array.tolist()
print(my_list)
#output
[1, 2, 4, 5, 6, 7, 4, 5, 6, 7]
About the speed!
import time
s_time = time.time()
print 'start'
a = range(100000000)
del a[:]
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 3.25
s_time = time.time()
print 'start'
a = range(100000000)
a = []
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 2.11
p=[2,3,4,4,4]
p.clear()
print(p)
[]
Only with Python 3
What's wrong with:
Motor=['1','2','2']
for i in Motor:
if i != '2':
print(i)
print(motor)