This question already has answers here:
Different ways of clearing lists
(8 answers)
Closed 4 years ago.
list1 = [2,5,61,7,10]
list1.remove(list1[0:len(list1)-1])
print(list1)
I want to remove all elements from that list but it shows me syntax error.
Any idea how can I remove all elements and print the final result like []
To remove all list items just use the in-built .clear() function:
>>> list1 = [2,5,61,7,10]
>>> list1.clear()
>>> print(list1)
[]
>>>
If you want to remove all elements from a list, you can use the slice assignment:
list1[:] = []
But with list.remove(), you can only do it one-by-one:
for item in list(list1):
list1.remove(item)
Note I created a copy of list1 with the for loop because it's dangerous to modify what you're iterating over, while it's safe to modify the list while iterating over a copy of it.
To remove some of the items:
for item in list1[0:3]:
list1.remove(item)
Or just better yet, use slice assignment:
list1[0:3] = []
Related
This question already has answers here:
Checking whether a string starts with XXXX
(5 answers)
Apply function to each element of a list
(4 answers)
Closed 3 months ago.
I have a list that I want to edit. For all the elements that start with '[x]', I want the list to remove those elements and place them into new list. But for the new list, in doing so, it should remove the '[x]' from the front of the elements.
list1 = ['[x]final', '[x]gym', 'midterm', '[x]class', 'school']
list1 becomes:
list1 = ['midterm', 'school']
new list that was created from removing the elements that had '[x]' in the front:
new_list = ['final', 'gym', 'class']
I am new to coding and am having difficulties with this.
Since you are a beginner, the verbose way to do this is the following
list1 = ['[x]final', '[x]gym', 'midterm', '[x]class', 'school']
new_list = []
for s in list1:
if s.startswith("[x]"):
new_list.append(s[3:])
print(new_list)
However, you can take advantage of Python's list comprehension to do it in a single line, like this
list1 = ['[x]final', '[x]gym', 'midterm', '[x]class', 'school']
new_list = [s[3:] for s in list1 if s[0:3] == "[x]"]
print(new_list)
Both methods yield ['final', 'gym', 'class']
This question already has answers here:
Checking if List contains all items from another list
(3 answers)
Closed 1 year ago.
list1 = ['Gryffindor', 'Ravenclaw', 'Hufflepuff', 'Slytherin']
list2 = ['Gryffindor', 'Ravenclaw']
checkif = item in List2 for item in List1
if check is True:
print("The list {} contains all elements of the list {}".format(List1, List2))
Why is this damn thing not working? Also is checkif = item in list2 for item in list1 a list comprehension or what?
Someone please correct my code, thanks.
I think you want all:
list1 = ['Gryffindor', 'Ravenclaw', 'Hufflepuff', 'Slytherin']
list2 = ['Gryffindor', 'Ravenclaw']
checkif = all(item in list1 for item in list2)
Also, you need to swap list1 and list1 to get the result you describe in the print line.
You can use sets here and specifically check the relationship whether one of your sets is a subset of the other:
set(list2).issubset(list1) # True
To use this the first object must be a set hence set(list2) but the second can be any iterable. One caveat here is that since we're comparing sets, it will only check for unique elements, i.e. it will not care about repeated values.
This question already has answers here:
Removing duplicates in lists
(56 answers)
Closed 5 years ago.
I have a list of strings in which there are a lot of repeated items. I would like to make a new list in which all items are present but there is only one occurrence of each item.
input:
mylist = ["hg", "yt", "hg", "tr", "yt"]
output:
newlist = ["hg", "yt", "tr"]
I actually have tried this code but did not return what I want:
newlist = []
for i in range(len(mylist)):
if mylist[i+1] == mylist[i]:
newlist.append(mylist[i])
You can simply use a set:
newlist = set(mylist)
Or, to retrieve exactly a list, but is can be useless depending what you are doing with:
nexlist = list(set(mylist))
This question already has answers here:
Ellipsis lists [...] and concatenating a list to itself [duplicate]
(3 answers)
Closed 6 years ago.
I have a list X = ['xyz']
I use the below commands for appending to another variable.
L = X
L.append(X)
L
Out[109]: ['xyz', [...]]
I am unable to understand why the second element in the new L list is not having the value as 'xyz'
My question is not how to append or extend a list, but in fact about the functionality of Append function, which has been correctly explained by #sinsuren below.
append add X as an element to L. If you want the element inside X to be inserted to L, use extend instead:
>>> X = ['xyz']
>>> L = X
>>> L.extend(X)
>>> L
['xyz', 'xyz']
Try this, it will extend the list.
L.extend(X)
But if you want to use append. Use a element like this L.append('abc'). It will give same result or L.append(X[0])
Edit: You have Appended list to itself. It will recursively append to itself and due to which even L[1] will give you same response. Like L[1] = ['xyz', [...]] . and for more understanding Please refer What's exactly happening in infinite nested lists?
This question already has answers here:
How to remove items from a list while iterating?
(25 answers)
Closed 7 years ago.
I am iterating through two lists like so:
list1 = [z]
list2 = [z-t,z-s,s-a,a-n, g-z]
for e in list1:
for t in list2:
# if item from list2 contains item from list1 (before "-")
# remove the item from list2 and append part after "-" to list1
# iterate again until no change is made or list2 is empty
The problem I can't solve is that when I remove the item from list2 it goes to next item and I am not able to prevent it. For example
list2 = [z-t,z-s,s-a....]
^ removing this while iterating
next iteration jumps over one item
list2 = [z-s, s-a,...]
^ this is where I am second iteration
list2 = [z-s, s-a,...]
^ this is where I want to be
Is there any way to do this?
There are a few ways to achieve this, one is iterating backwards over the list, ie
for e in list1:
for t in list2[::-1]: # This iterates over the list backwards
# Do your check
# Remove the element
If you need to process it forwards, you can iterate over a copy so you can mutate the original list while iterating over the initial content, ie
for e in list1:
for t in list2[:]: # The [:] syntax feeds you your data via a new list.
# Do your check
# Remove the element, and not worry about mutating the original list length