loop over list of list python

Question

I faced an issue with my code where the loop stops running once it removes the list from the list of list.

data=[["why","why","hello"],["why","why","bell"],["why","hi","sllo"],["why","cry","hello"]]

for word_set in data:
    if word_set[-1]!="hello":
        data.remove(word_set)
print(data)

My desired output is

[['why', 'why', 'hello'], ['why', 'cry', 'hello']]

but the output is

[['why', 'why', 'hello'], ['why', 'hi', 'sllo'], ['why', 'cry', 'hello']]

How do I make the loop go on till the end of the list?

sunhs · Accepted Answer · 2015-11-23 06:52:38Z

4

That's because, when you remove the second item (whose index is 1), the items after it move forward. In the next iteration, the index is 2. It should have been pointing to ["why","hi","solo"]. But since the items moved forward, it points to ["why","cry","hello"]. That's why you get the wrong result.

It's not recommended to remove list items while iterating over the list.

You can either create a new list (which is mentioned in the first answer) or use the filter function.

def filter_func(item):
    if item[-1] != "hello":
        return False
    return True

new_list = filter(filter_func, old_list)

answered Nov 23, 2015 at 6:52

sunhs

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LUCAS · Accepted Answer · 2015-11-23 07:38:16Z

2

Remember

data = [["list", "in","a list"],["list", "in","a list"],["list", "in","a list"]]
#data[0] will return ["list", "in","a list"]
#data[0][0] will return "list"
#remember that lists starts with '0' -> data[0]

answered Nov 23, 2015 at 7:38

LUCAS

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Mayur Koshti · Accepted Answer · 2015-11-23 06:47:38Z

1

>>> data=[["why","why","hello"],["why","why","bell"],["why","hi","sllo"],["why","cry","hello"]]
>>> y = []
>>> for subData in data:
        for dataItem in subData:
            if dataItem == "hello":
                y.append(subData)
>>> y
[['why', 'why', 'hello'], ['why', 'cry', 'hello']]

answered Nov 23, 2015 at 6:47

Mayur Koshti

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Comments

Akshay Hazari · Accepted Answer · 2015-11-23 07:28:11Z

1

filter(lambda x : x[-1] =="hello",[["why","why","hello"],["why","why","bell"],["why","hi","sllo"],["why","cry","hello"]])

OR

reduce(lambda x,y : x + [y] if y[-1]=="hello" else x ,[["why","why","hello"],["why","why","bell"],["why","hi","sllo"],["why","cry","hello"]],[])

OR

[i for i in [["why","why","hello"],["why","why","bell"],["why","hi","sllo"],["why","cry","hello"]] if i[-1]=="hello"]

edited Nov 23, 2015 at 7:28

answered Nov 23, 2015 at 7:15

Akshay Hazari

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Comments

John Zhang · Accepted Answer · 2015-11-23 09:49:55Z

1

data=[["why","why","hello"],["why","why","bell"],["why","hi","sllo"],["why","cry","hello"]]

for word_set in data[:]:
    if word_set[-1]!= "hello":
        data.remove(word_set)
print(data)

Don't iterate the origin data,but make a duplicate(data[:]). Because when remove items from list, the index of item will change.["why","why","bell"] in list index is 1. when it's removed from data. ["why","hi","sllo"] in data index will be 1. The next iteration index is 2, so ["why","hi","sllo"] is passed and checks ["why","cry","hello"].

edited Nov 23, 2015 at 9:49

answered Nov 23, 2015 at 8:48

John Zhang

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1 Comment

Tim Over a year ago

Please add some comment as to why you think this code is the answer, as it stands, this answer is not very useful.

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