从嵌套列表中的所有子列表中删除第一个元素

Question

I have a list of lists such that

a = [["append",5],["insert",7],["print",9,10]]

How can I remove the first value from each sublist within the list using map and lambda function?

This is my code

m =  map(lambda x : x.remove(x[0]), a)

And my output for list(m) is [None, None, None]

Can anybody please help me understand why this is happening and how can the code be corrected? Thanks!!!

Answer 1

If you really want to do the map way, here is the working version:

m = list(map(lambda x: x[1:], a)) 

# or even better just use List Comp.

m1 = [x[1:] for x in a]

assert m == m1  

>>> m
[[5], [7], [9, 10]]

Answer 2

Do you want to modify the existing lists or create new ones? Daniel Hao already covered creating new ones , so I'll cover modifying the existing ones.

You're getting None s because in Python, mutator methods return None , and list.remove() is one such mutator method. For more info, see Why does append() always return None in Python?

The better solution is to not use map at all, because you'd be using it primarily for side-effects, which is bad, the same way that using a list comprehension for side-effects is bad . Just use a plain for-loop:

for sublist in a:
    del sublist[0]

Also note that I'm using del here. Using remove() as you are is redundant: instead of simply saying, "delete the first element", you're saying "get the first element then find the first occurrence and remove it".

Answer 3

Here a functional approach to slice the items of each (sub)list.

Sometimes it is more the effort than the undertaking , see wjandrea 's answer.

from operator import itemgetter

a = [["append",5],["insert",7],["print",9,10]]


print(list(map(itemgetter(slice(1, None)), a)))

---

Here a dirty hack to fix the behavior of the approach in the question:

print(list(map(lambda x: (x.pop(0), x)[1], a)))

Also remove can be used instead pop .

Remember that also the original object will be affected by such operations!

Answer 4

It is happening because remove function is not returning anything whereas pop returns whatever element you remove from the list and instead of using "m" which is returned values of function remove use "a" which is modified list you can also use pop for it like this.

a = [["append",5],["insert",7],["print",9,10]]
m = list(map(lambda x : x.pop(0), a))

Than if you print m you will get

['append', 'insert', 'print']

And if you print a than you will get

[[5], [7], [9, 10]]