If I have a list like so:
List = ['a', 'b', 'c', 'd', 'e',]
what is the simplest way to get a Boolean answer if, I for instance asked it if 'g' was inside this list?
print 'g' in ['a', 'b', 'c', 'd']
l = ['a','b','c','d']
if 'g' in l:
print True
List = ['a', 'b', 'c', 'd', 'e']
def inlist (lst, character):
if character in lst and type(lst) is list:
return True
else:
return False
print inlist(List, 'g')
As you expect, this prints: False
NOTE: try to name your lists other than List, as that can cause some confusion when reading.
Related
I am trying to insert an element in the list at multiple instances. But by doing this, the length of the list is constantly changing. So, it is not reaching the last element.
my_list = ['a', 'b', 'c', 'd', 'e', 'a']
aq = len(my_list)
for i in range(aq):
if my_list[i] == 'a':
my_list.insert(i+1, 'g')
aq = aq+1
print(my_list)
The output I am getting is -
['a', 'g', 'b', 'c', 'd', 'e', 'a']
The output I am trying to get is -
['a', 'g', 'b', 'c', 'd', 'e', 'a', 'g']
How can I get that?
Changing aq in the loop does not change the range. That created an iterator when you entered the loop, and that iterator won't change. There are two ways to do this. The easy way is to build a new list:
newlist = []
for c in my_list:
newlist.append(c)
if c == 'a':
newlist.append('g')
The trickier way is to use .find() to find the next instance of 'a' and insert a 'g' after it, then keep searching for the next one.
Here is a nice way to write it using the built-in itertools.chain.from_iterable:
from itertools import chain
my_list = ['a', 'b', 'c', 'd', 'e', 'a']
my_list = list(chain.from_iterable((x, "g") if x == "a" else x for x in my_list))
# ['a', 'g', 'b', 'c', 'd', 'e', 'a', 'g']
Here, every occurance of "a" is replaced with "a", "g" in the list, otherwise the elements are left alone.
Say I have a list of options and I want to pick a certain number randomly.
In my case, say the options are in a list ['a', 'b', 'c', 'd', 'e'] and I want my script to return 3 elements.
However, there is also the case of two options that cannot appear at the same time. That is, if option 'a' is picked randomly, then option 'b' cannot be picked. And the same applies the other way round.
So valid outputs are: ['a', 'c', 'd'] or ['c', 'd', 'b'], while things like ['a', 'b', 'c'] would not because they contain both 'a' and 'b'.
To fulfil these requirements, I am fetching 3 options plus another one to compensate a possible discard. Then, I keep a set() with the mutually exclusive condition and keep removing from it and check if both elements have been picked or not:
import random
mutually_exclusive = set({'a', 'b'})
options = ['a', 'b', 'c', 'd', 'e']
num_options_to_return = 3
shuffled_options = random.sample(options, num_options_to_return + 1)
elements_returned = 0
for item in shuffled_options:
if elements_returned >= num_options_to_return:
break
if item in mutually_exclusive:
mutually_exclusive.remove(item)
if not mutually_exclusive:
# if both elements have appeared, then the set is empty so we cannot return the current value
continue
print(item)
elements_returned += 1
However, I may be overcoding and Python may have better ways to handle these requirements. Going through random's documentation I couldn't find ways to do this out of the box. Is there a better solution than my current one?
One way to do this is use itertools.combinations to create all of the possible results, filter out the invalid ones and make a random.choice from that:
>>> from itertools import combinations
>>> from random import choice
>>> def is_valid(t):
... return 'a' not in t or 'b' not in t
...
>>> choice([
... t
... for t in combinations('abcde', 3)
... if is_valid(t)
... ])
...
('c', 'd', 'e')
Maybe a bit naive, but you could generate samples until your condition is met:
import random
options = ['a', 'b', 'c', 'd', 'e']
num_options_to_return = 3
mutually_exclusive = set({'a', 'b'})
while True:
shuffled_options = random.sample(options, num_options_to_return)
if all (item not in mutually_exclusive for item in shuffled_options):
break
print(shuffled_options)
You can restructure your options.
import random
options = [('a', 'b'), 'c', 'd', 'e']
n_options = 3
selected_option = random.sample(options, n_options)
result = [item if not isinstance(item, tuple) else random.choice(item)
for item in selected_option]
print(result)
I would implement it with sets:
import random
mutually_exclusive = {'a', 'b'}
options = ['a', 'b', 'c', 'd', 'e']
num_options_to_return = 3
while True:
s = random.sample(options, num_options_to_return)
print('Sample is', s)
if not mutually_exclusive.issubset(s):
break
print('Discard!')
print('Final sample:', s)
Prints (for example):
Sample is ['a', 'b', 'd']
Discard!
Sample is ['b', 'a', 'd']
Discard!
Sample is ['e', 'a', 'c']
Final sample: ['e', 'a', 'c']
I created the below function and I think it's worth sharing it too ;-)
def random_picker(options, n, mutually_exclusives=None):
if mutually_exclusives is None:
return random.sample(options, n)
elif any(len(pair) != 2 for pair in mutually_exclusives):
raise ValueError('Lenght of pairs of mutually_exclusives iterable, must be 2')
res = []
while len(res) < n:
item_index = random.randint(0, len(options) - 1)
item = options[item_index]
if any(item == exc and pair[-(i - 1)] in res for pair in mutually_exclusives
for i, exc in enumerate(pair)):
continue
res.append(options.pop(item_index))
return res
Where:
options is the list of available options to pick from.
n is the number of items you want to be picked from options
mutually_exclusives is an iterable containing tuples pairs of mutually exclusive items
You can use it as follows:
>>> random_picker(['a', 'b', 'c', 'd', 'e'], 3)
['c', 'e', 'a']
>>> random_picker(['a', 'b', 'c', 'd', 'e'], 3, [('a', 'b')])
['d', 'b', 'e']
>>> random_picker(['a', 'b', 'c', 'd', 'e'], 3, [('a', 'b'), ('a', 'c')])
['e', 'd', 'a']
import random
l = [['a','b'], ['c'], ['d'], ['e']]
x = [random.choice(i) for i in random.sample(l,3)]
here l is a two-dimensional list, where the fist level reflects an and relation and the second level an or relation.
I am trying to understand the process of creating a function that can replace duplicate strings in a list of strings. for example, I want to convert this list
mylist = ['a', 'b', 'b', 'a', 'c', 'a']
to this
mylist = ['a', 'b', 'x', 'x', 'c', 'x']
initially, I know I need create my function and iterate through the list
def replace(foo):
newlist= []
for i in foo:
if foo[i] == foo[i+1]:
foo[i].replace('x')
return foo
However, I know there are two problems with this. the first is that I get an error stating
list indices must be integers or slices, not str
so I believe I should instead be operating on the range of this list, but I'm not sure how to implement it. The other being that this would only help me if the duplicate letter comes directly after my iteration (i).
Unfortunately, that's as far as my understanding of the problem reaches. If anyone can provide some clarification on this procedure for me, I would be very grateful.
Go through the list, and keep track of what you've seen in a set. Replace things you've seen before in the list with 'x':
mylist = ['a', 'b', 'b', 'a', 'c', 'a']
seen = set()
for i, e in enumerate(mylist):
if e in seen:
mylist[i] = 'x'
else:
seen.add(e)
print(mylist)
# ['a', 'b', 'x', 'x', 'c', 'x']
Simple Solution.
my_list = ['a', 'b', 'b', 'a', 'c', 'a']
new_list = []
for i in range(len(my_list)):
if my_list[i] in new_list:
new_list.append('x')
else:
new_list.append(my_list[i])
print(my_list)
print(new_list)
# output
#['a', 'b', 'b', 'a', 'c', 'a']
#['a', 'b', 'x', 'x', 'c', 'x']
The other solutions use indexing, which isn't necessarily required.
Really simply, you could check if the value is in the new list, else you can append x. If you wanted to use a function:
old = ['a', 'b', 'b', 'a', 'c']
def replace_dupes_with_x(l):
tmp = list()
for char in l:
if char in tmp:
tmp.append('x')
else:
tmp.append(char)
return tmp
new = replace_dupes_with_x(old)
You can use the following solution:
from collections import defaultdict
mylist = ['a', 'b', 'b', 'a', 'c', 'a']
ret, appear = [], defaultdict(int)
for c in mylist:
appear[c] += 1
ret.append(c if appear[c] == 1 else 'x')
Which will give you:
['a', 'b', 'x', 'x', 'c', 'x']
I create a list, and I want to remove a string from it.
Ex:
>>> myList = ['a', 'b', 'c', 'd']
>>> myList = myList.remove('c')
>>> print(myList)
None
What am I doing wrong here? All I want is 'c' to be removed from myList!
I am not sure what a is (I am guessing another list), you should do myList.remove() alone, without assignment.
Example -
>>> myList = ['a', 'b', 'c', 'd']
>>> myList.remove('c')
>>> myList
['a', 'b', 'd']
myList.remove() does not return anything, hence when you do myList = <anotherList>.remove(<something>) it sets myList to None
Remember that lists are mutable, so you can simply call remove on the list itself:
>>> myList = ['a', 'b', 'c', 'd']
>>> myList.remove('c')
>>> myList
['a', 'b', 'd']
The reason you were getting None before is because remove() always returns None
Just an addition to Anand's Answer,
mylist = mylist.remove('c')
The above code will return 'none' as the return type for my list. So you want to keep it as
mylist.remove('c')
The remove() function doesn't return anything, it modifies the list in place. If you don't assign it to a variable you will see that myList doesn't contain c anymore.
This question already has answers here:
Loop "Forgets" to Remove Some Items [duplicate]
(10 answers)
Closed 8 years ago.
I have a simple question about lists
Suppose that I want to delete all 'a's from a list:
list = ['a', 'a', 'b', 'b', 'c', 'c']
for element in list:
if element == 'a':
list.remove('a')
print list
==> result:
['a', 'b', 'b', 'c', 'c', 'd', 'd']
I know this is happening because, after I remove the first 'a', the list index gets
incremented while all the elements get pushed left by 1.
In other languages, I guess one way to solve this is to iterate backwards from the end of the list..
However, iterating through reversed(list) returns the same error.
Is there a pythonic way to solve this problem??
Thanks
One of the more Pythonic ways:
>>> filter(lambda x: x != 'a', ['a', 'a', 'b', 'b', 'c', 'c'])
['b', 'b', 'c', 'c']
You should never modify a list while iterating over it.
A better approach would be to use a list comprehension to exclude an item:
list1 = ['a', 'a', 'b', 'b', 'c', 'c']
list2 = [x for x in list1 if x != 'a']
Note: Don't use list as a variable name in Python - it masks the built-in list type.
You are correct, when you remove an item from a list while iterating over it, the list index gets out of sync. What both the other existing answers are hinting at is that you need to create a new list and copy over only the items you want.
For example:
existing_list = ['a', 'a', 'b', 'c', 'd', 'e']
new_list = []
for element in existing_list:
if element != 'a':
new_list.append(element)
existing_list = new_list
print existing_list
outputs: ['b', 'c', 'd', 'e']