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remove data from old list and append in new list python [duplicate]

This question already has answers here:
How to remove items from a list while iterating?
(25 answers)
Closed 7 years ago.
I have a list
a = ["a", "b", "c", "d", "e"]
I want to remove elements in this list in a for loop like below:
for item in a:
print(item)
a.remove(item)
But it doesn't work. What can I do?

You are not permitted to remove elements from the list while iterating over it using a for loop.
The best way to rewrite the code depends on what it is you're trying to do.
For example, your code is equivalent to:
for item in a:
print(item)
a[:] = []
Alternatively, you could use a while loop:
while a:
print(a.pop())
I'm trying to remove items if they match a condition. Then I go to next item.
You could copy every element that doesn't match the condition into a second list:
result = []
for item in a:
if condition is False:
result.append(item)
a = result
Alternatively, you could use filter or a list comprehension and assign the result back to a:
a = filter(lambda item:... , a)
or
a = [item for item in a if ...]
where ... stands for the condition that you need to check.

Iterate through a copy of the list:
>>> a = ["a", "b", "c", "d", "e"]
>>> for item in a[:]:
print(item)
if item == "b":
a.remove(item)
a
b
c
d
e
>>> print(a)
['a', 'c', 'd', 'e']

As other answers have said, the best way to do this involves making a new list - either iterate over a copy, or construct a list with only the elements you want and assign it back to the same variable. The difference between these depends on your use case, since they affect other variables for the original list differently (or, rather, the first affects them, the second doesn't).
If a copy isn't an option for some reason, you do have one other option that relies on an understanding of why modifying a list you're iterating breaks. List iteration works by keeping track of an index, incrementing it each time around the loop until it falls off the end of the list. So, if you remove at (or before) the current index, everything from that point until the end shifts one spot to the left. But the iterator doesn't know about this, and effectively skips the next element since it is now at the current index rather than the next one. However, removing things that are after the current index doesn't affect things.
This implies that if you iterate the list back to front, if you remove an item at the current index, everything to it's right shifts left - but that doesn't matter, since you've already dealt with all the elements to the right of the current position, and you're moving left - the next element to the left is unaffected by the change, and so the iterator gives you the element you expect.
TL;DR:
>>> a = list(range(5))
>>> for b in reversed(a):
if b == 3:
a.remove(b)
>>> a
[0, 1, 2, 4]
However, making a copy is usually better in terms of making your code easy to read. I only mention this possibility for sake of completeness.

import copy
a = ["a", "b", "c", "d", "e"]
b = copy.copy(a)
for item in a:
print(item)
b.remove(item)
a = copy.copy(b)
Works: to avoid changing the list you are iterating on, you make a copy of a, iterate over it and remove the items from b. Then you copy b (the altered copy) back to a.

How about creating a new list and adding elements you want to that new list. You cannot remove elements while iterating through a list

Probably a bit late to answer this but I just found this thread and I had created my own code for it previously...
list = [1,2,3,4,5]
deleteList = []
processNo = 0
for item in list:
if condition:
print(item)
deleteList.insert(0, processNo)
processNo += 1
if len(deleteList) > 0:
for item in deleteList:
del list[item]
It may be a long way of doing it but seems to work well. I create a second list that only holds numbers that relate to the list item to delete. Note the "insert" inserts the list item number at position 0 and pushes the remainder along so when deleting the items, the list is deleted from the highest number back to the lowest number so the list stays in sequence.

Removing dups FOR loop python

I am trying to write a function to remove duplicates in list using a for loop in python. This is what I've got:
[IN]:
L = [1, 1, "a", "b", 1, "a", "d"]
N = [ ]
def remove_dup(x):
for x in range(len(L)):
if L[x] not in N:
N.append(x)
return N
print(remove_dup(N))
My problem is instead of outputting [1, "a", "b", "d"] its returning this:
[OUT]:
[1, 2, 3, 4, 5, 6]
Lol idk where there is coming from. At least its printing a list though.... Can someone help a girl out and tell me why I am getting a randomly generated list? I have tried switching out a bunch of variables when calling my function.
Thank you!

To N, you are appending x instead of L[x].

You're iterating over the indices rather than the values, and then appending those to your new list. You need to either look up the element to add it, or just iterate over the values directly -- since you don't actually need the index for anything, this is going to be more convenient and readable code.
Separately, it's not a good idea to use a global list like that to return results, since it'll cause problems when you try to use your function more than once. Instead, create the "return" list inside your function. Also, you're not actually using the list passed in to the function (named x) -- you're just looking up values in the global L list instead. Which will also cause problems if you try to use your function with a different list.
So to fix all of those issues, and with variable names edited for readability:
# Iterating over indices
def remove_dups(in_list):
result = []
for index in range(len(in_list)):
if in_list[index] not in result:
result.append(in_list[index])
return result
# Or over values (more readable):
def remove_dups(in_list):
result = []
for item in in_list:
if item not in result:
result.append(item)
return result
Finally - using in on a list isn't the best way to check for duplicates, since it'll search the whole list every time. A much more convenient and performant way of doing this is to just convert it to a set and back:
def remove_dups(in_list):
return list(set(in_list))
With the caveat that this will not keep the list elements in their original order.

Instead of N.append(x) use:
N.append(L[x])
In your case x is an index that goes from 0 to the size of L minus one. But you need the values that can be taken from L by index x.

In addition to the other answers about using L[x] as opposed to just x you could do something like this instead:
L = [1, 1, "a", "b", 1, "a", "d"]
N = [ ]
def remove_dup(x):
for x in L:
if x not in N:
N.append(x)
return N
print(remove_dup(N))
This way is slightly more readable in my opinion, instead of using indexes you're using the values. Either way works.

Iterating over a list / selecting data type for removal from source list with for loop [duplicate]

This question already has answers here:
How to remove items from a list while iterating?
(25 answers)
Closed 7 years ago.
I have a list
a = ["a", "b", "c", "d", "e"]
I want to remove elements in this list in a for loop like below:
for item in a:
print(item)
a.remove(item)
But it doesn't work. What can I do?

You are not permitted to remove elements from the list while iterating over it using a for loop.
The best way to rewrite the code depends on what it is you're trying to do.
For example, your code is equivalent to:
for item in a:
print(item)
a[:] = []
Alternatively, you could use a while loop:
while a:
print(a.pop())
I'm trying to remove items if they match a condition. Then I go to next item.
You could copy every element that doesn't match the condition into a second list:
result = []
for item in a:
if condition is False:
result.append(item)
a = result
Alternatively, you could use filter or a list comprehension and assign the result back to a:
a = filter(lambda item:... , a)
or
a = [item for item in a if ...]
where ... stands for the condition that you need to check.

Iterate through a copy of the list:
>>> a = ["a", "b", "c", "d", "e"]
>>> for item in a[:]:
print(item)
if item == "b":
a.remove(item)
a
b
c
d
e
>>> print(a)
['a', 'c', 'd', 'e']

As other answers have said, the best way to do this involves making a new list - either iterate over a copy, or construct a list with only the elements you want and assign it back to the same variable. The difference between these depends on your use case, since they affect other variables for the original list differently (or, rather, the first affects them, the second doesn't).
If a copy isn't an option for some reason, you do have one other option that relies on an understanding of why modifying a list you're iterating breaks. List iteration works by keeping track of an index, incrementing it each time around the loop until it falls off the end of the list. So, if you remove at (or before) the current index, everything from that point until the end shifts one spot to the left. But the iterator doesn't know about this, and effectively skips the next element since it is now at the current index rather than the next one. However, removing things that are after the current index doesn't affect things.
This implies that if you iterate the list back to front, if you remove an item at the current index, everything to it's right shifts left - but that doesn't matter, since you've already dealt with all the elements to the right of the current position, and you're moving left - the next element to the left is unaffected by the change, and so the iterator gives you the element you expect.
TL;DR:
>>> a = list(range(5))
>>> for b in reversed(a):
if b == 3:
a.remove(b)
>>> a
[0, 1, 2, 4]
However, making a copy is usually better in terms of making your code easy to read. I only mention this possibility for sake of completeness.

import copy
a = ["a", "b", "c", "d", "e"]
b = copy.copy(a)
for item in a:
print(item)
b.remove(item)
a = copy.copy(b)
Works: to avoid changing the list you are iterating on, you make a copy of a, iterate over it and remove the items from b. Then you copy b (the altered copy) back to a.

How about creating a new list and adding elements you want to that new list. You cannot remove elements while iterating through a list

Probably a bit late to answer this but I just found this thread and I had created my own code for it previously...
list = [1,2,3,4,5]
deleteList = []
processNo = 0
for item in list:
if condition:
print(item)
deleteList.insert(0, processNo)
processNo += 1
if len(deleteList) > 0:
for item in deleteList:
del list[item]
It may be a long way of doing it but seems to work well. I create a second list that only holds numbers that relate to the list item to delete. Note the "insert" inserts the list item number at position 0 and pushes the remainder along so when deleting the items, the list is deleted from the highest number back to the lowest number so the list stays in sequence.

How to remove list elements in a for loop in Python? [duplicate]

This question already has answers here:
How to remove items from a list while iterating?
(25 answers)
Closed 7 years ago.
I have a list
a = ["a", "b", "c", "d", "e"]
I want to remove elements in this list in a for loop like below:
for item in a:
print(item)
a.remove(item)
But it doesn't work. What can I do?

You are not permitted to remove elements from the list while iterating over it using a for loop.
The best way to rewrite the code depends on what it is you're trying to do.
For example, your code is equivalent to:
for item in a:
print(item)
a[:] = []
Alternatively, you could use a while loop:
while a:
print(a.pop())
I'm trying to remove items if they match a condition. Then I go to next item.
You could copy every element that doesn't match the condition into a second list:
result = []
for item in a:
if condition is False:
result.append(item)
a = result
Alternatively, you could use filter or a list comprehension and assign the result back to a:
a = filter(lambda item:... , a)
or
a = [item for item in a if ...]
where ... stands for the condition that you need to check.

Iterate through a copy of the list:
>>> a = ["a", "b", "c", "d", "e"]
>>> for item in a[:]:
print(item)
if item == "b":
a.remove(item)
a
b
c
d
e
>>> print(a)
['a', 'c', 'd', 'e']

As other answers have said, the best way to do this involves making a new list - either iterate over a copy, or construct a list with only the elements you want and assign it back to the same variable. The difference between these depends on your use case, since they affect other variables for the original list differently (or, rather, the first affects them, the second doesn't).
If a copy isn't an option for some reason, you do have one other option that relies on an understanding of why modifying a list you're iterating breaks. List iteration works by keeping track of an index, incrementing it each time around the loop until it falls off the end of the list. So, if you remove at (or before) the current index, everything from that point until the end shifts one spot to the left. But the iterator doesn't know about this, and effectively skips the next element since it is now at the current index rather than the next one. However, removing things that are after the current index doesn't affect things.
This implies that if you iterate the list back to front, if you remove an item at the current index, everything to it's right shifts left - but that doesn't matter, since you've already dealt with all the elements to the right of the current position, and you're moving left - the next element to the left is unaffected by the change, and so the iterator gives you the element you expect.
TL;DR:
>>> a = list(range(5))
>>> for b in reversed(a):
if b == 3:
a.remove(b)
>>> a
[0, 1, 2, 4]
However, making a copy is usually better in terms of making your code easy to read. I only mention this possibility for sake of completeness.

import copy
a = ["a", "b", "c", "d", "e"]
b = copy.copy(a)
for item in a:
print(item)
b.remove(item)
a = copy.copy(b)
Works: to avoid changing the list you are iterating on, you make a copy of a, iterate over it and remove the items from b. Then you copy b (the altered copy) back to a.

How about creating a new list and adding elements you want to that new list. You cannot remove elements while iterating through a list

Probably a bit late to answer this but I just found this thread and I had created my own code for it previously...
list = [1,2,3,4,5]
deleteList = []
processNo = 0
for item in list:
if condition:
print(item)
deleteList.insert(0, processNo)
processNo += 1
if len(deleteList) > 0:
for item in deleteList:
del list[item]
It may be a long way of doing it but seems to work well. I create a second list that only holds numbers that relate to the list item to delete. Note the "insert" inserts the list item number at position 0 and pushes the remainder along so when deleting the items, the list is deleted from the highest number back to the lowest number so the list stays in sequence.

Removing items from a list in a loop [duplicate]

This question already has answers here:
Strange result when removing item from a list while iterating over it
(8 answers)
Closed 7 years ago.
For quite a bit of time now I have been trying to figure out a way to loop through a list and remove the current item that I'm at. I can't seem to get this working as I would like it to. It loops just 1 time through, but I wanted it to loop 2 times. When I remove the removal line - it loops 2 times.
a = [0, 1]
for i in a:
z = a
print z.remove(i)
The output:
[1]
The output that I was expecting:
[1]
[0]

You're changing the list while iterating over it -- z = a doesn't make a copy, it just points z at the same place a points.
Try
for i in a[:]: # slicing a list makes a copy
print i # remove doesn't return the item so print it here
a.remove(i) # remove the item from the original list
or
while a: # while the list is not empty
print a.pop(0) # remove the first item from the list
If you don't need an explicit loop, you can remove items that match a condition with a list comprehension:
a = [i for i in a if i] # remove all items that evaluate to false
a = [i for i in a if condition(i)] # remove items where the condition is False

It is bad practice modify a list while you're looping through it†. Create a copy of the list:
oldlist = ['a', 'b', 'spam', 'c']
newlist = [x for x in oldlist if x != 'spam']
To modify the original list, write the copy back in-place with a slice assignment:
oldlist[:] = [x for x in oldlist if x != 'spam']
† For a gist of why this might be bad practice, consider the implementation details of what goes on with the iterator over the sequence when the sequence changes during iteration. If you've removed the current item, should the iterator point to the next item in the original list or to the next item in the modified list? What if your decision procedure instead removes the previous (or next) item to the current?

The problem is that you're modifying a with remove so the loop exits because the index is now past the end of it.

Don't try to remove multiple items of a list while looping the list. I think it's a general rule you should follow not only in python but also in other programming languages as well.
You could add the item to be removed into a separate list. And then remove all objects in that new list from the original list.