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I believe this is an easy problem to solve. I have searched and found a few similar answers but not an efficient way to exactly what I want to achieve.
Assuming the following list:
x = [6, 7, 8]
I want to create a new list by repeating each number k times. Assuming k=3, the result should be:
xr = [6, 6, 6, 7, 7, 7, 8, 8, 8]
I was able to accomplish this using nest loops, which I believe is very inefficient:
xr = []
for num in x: # for each number in the list
for t in range(3): # repeat 3 times
xx2.append(num)
I also tried:
[list(itertools.repeat(x[i], 3)) for i in range(len(x))]
but I get:
[[6, 6, 6], [7, 7, 7], [8, 8, 8]]
Is there a more efficient direct method to accomplish this?
You can use list comprehension:
x = [6, 7, 8]
k = 3
out = [v for v in x for _ in range(k)]
print(out)
Prints:
[6, 6, 6, 7, 7, 7, 8, 8, 8]
def repeat_k(l,k):
lo = []
for x in l:
for i in range(k):
lo.append(x)
return lo
print (repeat_k([1,2,3],5))
Output:
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3]
With list comprehension:
def repeat_k(l,k):
return [ x for x in l for i in range(k) ]
print (repeat_k([1,2,3],5))
Output:
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3]
Another possibility:
>>> x = [6, 7, 8]
>>> k = 3
>>> l = []
>>> for item in x:
... l += k * [item]
...
>>> l
[6, 6, 6, 7, 7, 7, 8, 8, 8]
You can create a convenient function:
def repeat(it, n):
for elem in it: yield from [elem] * n
Use it like:
>>> list(repeat(x, n=3))
[6, 6, 6, 7, 7, 7, 8, 8, 8]
Thanks, everyone for the answers.
It seems there is an easier and more direct way to solve this using Numpy.
np.repeat(x, 3).tolist()
prints exactly what I needed:
[6, 6, 6, 7, 7, 7, 8, 8, 8]
import itertools
x=[4,5,6]
k=3
res = list(itertools.chain.from_iterable(itertools.repeat(i, K) for i in test_list))
print (res)
It can also be solved using python inbuilt functions of itertools library. The repeat function does the task of repetition and grouping into a list is done by the from_iterable function.
I have a list of integers. Each number can appear several times, the list is unordered.
I want to get the list of relative sizes. Meaning, if for example the original list is [2, 5, 7, 7, 3, 10] then the desired output is [0, 2, 3, 3, 1, 4]
Because 2 is the zero'th smallest number in the original list, 3 is one'th, etc.
Any clear easy way to do this?
Try a list comprehension with dictionary and also use set for getting unique values, like below:
>>> lst = [2, 5, 7, 7, 3, 10]
>>> newl = dict(zip(range(len(set(lst))), sorted(set(lst))))
>>> [newl[i] for i in lst]
[0, 2, 3, 3, 1, 4]
>>>
Or use index:
>>> lst = [2, 5, 7, 7, 3, 10]
>>> newl = sorted(set(lst))
>>> [newl.index(i) for i in lst]
[0, 2, 3, 3, 1, 4]
>>>
def list_move_back(new_value, value_list):
for i in reversed(value_list):
if value_list.index(i) != len(value_list)-1:
value_list[value_list.index(i)+1] = i
value_list[0] = new_value
return value_list
I want to get the following result:
list_example = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
list_example = list_move_back(12, list_example]
print(list_example)
>>>[12, 1, 2, 3, 4, 5, 6, 7, 8, 9]
It works if I run the function two times:
list_example = list_move_back(12, list_example]
print(list_example)
>>>[12, 12, 1, 2, 3, 4, 5, 6, 7, 8]
but if I want to run it a third time, the result looks like that:
list_example = list_move_back(12, list_example]
print(list_example)
>>>[12, 12, 1, 1, 3, 4, 5, 6, 7, 8]
The first 1 should be a 12. I have no idea why it doesn't work.
Just use list slicing:
def list_move_back(new_value, list_of_values):
return [new_value] + list_of_values[:-1]
Explanation: list_of_values[:-1] returns all the elements except for the last. By appending it to the new value, you get the wanted result. This answer has a pretty cool explanation of how list slicing works.
Also, if for some reason you'd like the "verbose" way to do this (maybe for an exercise or whatever), here's a way to go about it:
def list_move_back(new_value, list_of_values):
for i in range(len(list_of_values)-1, 0, -1):
list_of_values[i] = list_of_values[i-1]
list_of_values[0] = new_value
return list_of_values
I'd recommend list slicing over this method 9/10 times but again, I'm just leaving this here because there might be a case where someone wants to do this as some sort of mental exercise for indexing.
If you need the list to change in place, you can use the list methods .pop() to remove the last item and .insert(0,value) to add an item to the front:
>>> L = list(range(1,11))
>>> L
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> id(L)
1772071032392
>>> L.pop();L.insert(0,12)
10
>>> L
[12, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> id(L) # same list id, modified in place...
1772071032392
I intend to get the n smallest numbers in a list but keep the numbers in the same order they appear in the list. For example:
This is my list:
A = [1, 3, 4, 6, 7, 6, 8, 7, 2, 6, 8, 7, 0]
I like to get the first three lowest numbers as it has been ordered in the first list:
[1, 2, 0]
I do not want to sort the result as:
[0, 1, 2]
I have tried:
heapq.nsmallest(3,A)
but i wonder if it is possible to retain this list as:[1, 2, 0]
By the way, I'm not a Python coder so thanks for the help in advance.
You can try this:
new_a = []
A=[1, 3, 4, 6, 7, 6, 8, 7, 2, 6, 8, 7, 0]
for a in A:
if a not in new_a:
new_a.append(a)
new_a = [i for i in new_a if i in sorted(new_a)[:3]]
Output:
[1, 2, 0]
You could use heapq.nsmallest() to get the n smallest elements from the list. Then use collections.Counter to create a multiset from that list which you can use to check which elements from the original list to include in the result, e.g.
>>> from heapq import nsmallest
>>> from collections import Counter
>>> A = [1, 3, 4, 6, 7, 6, 8, 7, 2, 6, 8, 7, 0]
>>> n = 3
>>> c = Counter(nsmallest(n, A))
>>> result = []
>>> for elem in A:
... if c.get(elem, 0):
... result.append(elem)
... c[elem] -= 1
...
>>> result
[1, 2, 0]
Write a function or program that will take 2 arrays of integers, "current" and "target", and produce 2 arrays representing an additions list and a deletions list such that applying the additions and deletions to the "current" array will yield the "target" array.
For example, given the following
inputs:
current = [1, 3, 5, 6, 8, 9]
target = [1, 2, 5, 7, 9]
The outputs would be:
additions: [2, 7]
deletions: [3, 6, 8]
So that the following is true:
current([1, 3, 5, 6, 8, 9]) + additions([2, 7]) - deletions([3, 6, 8]) = target([1, 2, 5, 7, 9])
Solution:
So far I have this:
---------------------------
# import array function
from array import array
# create an integer array named current
current = array('i', [1, 3, 5, 6, 8, 9])
# add items from additions list into current array using the fromlist() method
additions = [2, 7]
current.fromlist(additions)
# remove items on deletions list from current array using the. remove() method
current.remove(3)
current.remove(6)
current.remove(8)
It will work for you...
def addlist(current,target):
add = []
intersection = set(current) & set(target)
for i in target:
if i not in intersection:
add.append(i)
return add
def removeList(current,target):
remove = []
intersection = set(current) & set(target)
for i in current:
if i not in intersection:
remove.append(i)
return remove
def main():
current = [1, 3, 5, 6, 8, 9]
target = [1, 2, 5, 7, 9]
print(addlist(current,target))
print(removeList(current,target))
if __name__=="__main__":
main()
Below one should be easy to understand.
>>> current = [1, 3, 5, 6, 8, 9]
>>> target = [1, 2, 5, 7, 9]
>>> set(current) & set(target)
set([1, 5, 9])
>>> unique = list(set(current) & set(target))
>>> additions = [i for i in target if i not in unique]
>>> additions
[2, 7]
>>> deletions = [i for i in current if i not in unique]
>>> deletions
[3, 6, 8]
>>>
This will work as well.
current = [1, 3, 5, 6, 8, 9]
target = [1, 2, 5, 7, 9]
additions=[x for x in target if x not in current]
deletions=[x for x in current if x not in target]