CODEWITHSUNDEEP
python
lte__
lte__

Reputation: 7576

mapping the append function to a list

I want to do the following elegantly. I have a list:

list1 = [[1,2],[3,1,4,7],[5],[7,8]]

I'd like to append the number 1 to each element of the list, so that I have

list1 = [[1,2,1],[3,1,4,7,1],[5,1],[7,8,1]]

I'm trying to map this via 

map(list.append([1]), vectors)

but this returns the error append() takes exactly one argument (0 given) and if I just try append([1]) (without list.), I get NameError: global name 'append' is not defined. I guess I could do it with a loop, but this seems more elegant, is there a way to map this correctly?

Upvotes: 3

Views: 6319

Answers (5)

Reishin
Reishin

Reputation: 1954

Here is a several ways to implement what you want:

More readable and classic way

for el in list1:
  el.append(1)

List comprehension

list1 = [el + [1] for el in list1]

Generators:

list1 = (el + [1] for el in list1)

Map

list1 = map(lambda el: el + [1], list1)

What to use?

It depends on you own situation and may depends on execution speed optimizations, code readability, place of usage.

  • Map is a worst choice in case of readability and execution speed
  • For is a fastest and more plain way to do this
  • Generators allows you to generate new list only when you really need this
  • List comprehension - one liner for classic for and it takes advantage when you need quickly filter the new list using if

i.e. if you need only add element to each item - for loop is a best choice to do this, but if you need add item only if item > 40, then you may consider to use List comprehension.

For example:

Classic For

x = 41

for el in list1:
  if x > 40:
    el.append(x)

List comprehension

   x = 1
   list1 = [el + [x] for el in list1 if x > 40]

as @jmd_dk mentioned, in this sample is one fundamental difference: with simple for you can just append to an existing object of the list which makes much less impact to execution time and memory usage. When you use List comprehension, you will get new list object and in this case new list object for each item.

Upvotes: 4

ettanany
ettanany

Reputation: 19806

With list comprehension and append, you can do:

list1 = [[1, 2], [3, 1, 4, 7], [5], [7, 8]]

[item.append(1) for item in list1]
print(list1)  # Output: [[1, 2, 1], [3, 1, 4, 7, 1], [5, 1], [7, 8, 1]]

Output:

>>> list1 = [[1, 2], [3, 1, 4, 7], [5], [7, 8]]
>>> [item.append(1) for item in list1]
[None, None, None, None]
>>> list1
[[1, 2, 1], [3, 1, 4, 7, 1], [5, 1], [7, 8, 1]]

You may also use extend like this:

[item.extend([1]) for item in list1]
print(list1)  # Output: [[1, 2, 1], [3, 1, 4, 7, 1], [5, 1], [7, 8, 1]]

Upvotes: 0

Zhang Tong
Zhang Tong

Reputation: 4719

map(lambda x: x + [1], list1)

you mean this?

list.append() have NO return (mean always return None)

Upvotes: 0

jmd_dk
jmd_dk

Reputation: 13090

You can simply do

list1 = [[1,2],[3,1,4,7],[5],[7,8]]
for el in list1:
    el.append(1)

Upvotes: 0

Exelian
Exelian

Reputation: 5888

Try a list comprehension, taking advantage of the fact the adding lists concats them together.

new_list = [l + [1] for l in list1]

Upvotes: 0

Related Questions