Lists and lambda expression

Question

First of all, hello, I have a little question. I'm trying to learn how to use lambda expression etc... I have this bit of code, it's an example that I need to replicate.

>>>something (lambda x:x+1, lambda y:y+10, [1, 2, 3, 4])
[2, 12, 4, 14]

I need that output and so far I got this:

l = [1,2,3,4]
def result(l):
    o = l[0::2]
    o2 = l[1::2]
    p = map(lambda x:x+1,o)
    p2 = map(lambda y:y+10,o2)
    return p,p2

First of all I know I'm returning 2 separated lists, I'm trying to figure that one out.

Is there any way to do this without having p and p2 separated? Something like this:

p = map(lambda x,y: x+1 x+10, o,o2)

I know that line doesn't work I'm just trying to illustrate what I'm asking

What is something?

jwodder
– jwodder

2016-06-15 18:23:40 +00:00
Commented Jun 15, 2016 at 18:23 — jwodder
– jwodder, Commented Jun 15, 2016 at 18:23
random name couldn't come up with a better one.

EXT
– EXT

2016-06-15 19:06:57 +00:00
Commented Jun 15, 2016 at 19:06 — EXT
– EXT, Commented Jun 15, 2016 at 19:06

Moses Koledoye · Accepted Answer · 2016-06-15 18:28:22Z

1

You're using those lambda functions wrongly. Why use multiple lambda functions to process a simple list? You can use just one, with a list comprehension:

r = [1, 2, 3, 4]
func = lambda x: [v+1 if i%2==0 else v+10 for i,v in enumerate(x)]
func(r)
# [2, 12, 4, 14]

You may even prefer writing a proper function for this for readability.

Or better, you can even do without the lambda function altogether:

[v+1 if i%2 == 0 else v+10 for i,v in enumerate(x)]

edited Jun 15, 2016 at 18:28

answered Jun 15, 2016 at 18:25

Moses Koledoye

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4 Comments

Two-Bit Alchemist Over a year ago

Simpler: [value + (1 if i % 2 == 0 else 10) for i, value in enumerate(lst)]

azalea Over a year ago

With a list comprehension, there is no need for a lambda function.

Moses Koledoye Over a year ago

@azalea OP apparently wants to understand more about lambda functions

Two-Bit Alchemist Over a year ago

Sounds like they should understand first that one's not needed here :D

M.T · Accepted Answer · 2016-06-15 19:13:48Z

0

Assuming each of the lambda expressions are used on every second value of the third argument to something you could simply overwrite the list(or alternatively make a new one):

def something(f1,f2,lst):
    for i,li in enumerate(lst):
        if not i%2: 
            lst[i] = f1(li)
        else:  
            lst[i] = f2(li)
    return lst

something (lambda x:x+1, lambda y:y+10, [1, 2, 3, 4])
#[2, 12, 4, 14]

This is by no means optimized, but it is readable.

answered Jun 15, 2016 at 19:13

M.T

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1 Comment

Two-Bit Alchemist Over a year ago

Well... that is something

jsbueno · Accepted Answer · 2016-06-16 05:02:49Z

0

The other answers will work, but if your question is just one case were you need to cycle several functions to be applied to different elements of a list, and then repeat the functions - then, semantically, a combination of zip and itertools.cycle would be a nice approach:

from itertools import cycle
[f(n) for f, n in zip(cycle((lambda x: x + 1, lambda x: x + 10)), [1, 2, 3, 4])]

Output:

[2, 12, 4, 14]

answered Jun 16, 2016 at 5:02

jsbueno

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