Constructing a list from another list in python

Question

I need to write a function of all_gt(nums, n) which nums is a list of number, n is a number, returns the list of numbers from nums that are greater than n. The order of elements should be preserved.

What I need to do is write in .append format and start by initializing the result to an empty list.

For example:

all_gt([1,2,3,4],4) => [] 
all_gt([2,3,4,5], 3) => [4,5]

This is what I have:

def all_gt(nums, n):
    for i in nums:
        if i > n:
            return nums.append(i)

I know what I did is wrong and I hope someone can help me solve this question.

You are on the right track. First, create a new list in the method. Then append all the numbers i that are greater than n to that new list. After the for loop ran, return the new list. — Stefan van den Akker
– Stefan van den Akker, Commented Mar 29, 2015 at 11:57
There are 3 or 4 issues here. First, append returns None and modifies nums. Second, you need to construct a new list and this is absent from your code. Third, you need to return once all the required elements are added to the list, in your code you will return after one element is added to your list. Fourth you should look at idiomatic ways of iterating over lists in Python using list comprehensions, even if your assignment requires you to use append. — YXD
– YXD, Commented Mar 29, 2015 at 11:57

Bhargav Rao · Accepted Answer · 2015-03-29 11:54:43Z

2

Your mistake

return nums.append(i) appends a value to the list in-place and returns None. A workaround can be

def all_gt(nums, n):
    num2 = []
    for i in nums:
        if i > n:
            num2.append(i)
    return num2

Better Ways

You can use a list comprehension

def all_gt(nums, n):
    return [i for i in nums if i > n]

answered Mar 29, 2015 at 11:54

Bhargav Rao

52.6k29 gold badges130 silver badges142 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

icefish Over a year ago

The return num2 return nothing.

Bhargav Rao Over a year ago

@icefish I have tested it. It works in my case. Just check if you have missed any part of my answer. Especially num2 = [] statement after the function definition

icefish Over a year ago

Oh my mistake, I still put a return infront of num2.append(i). It works now. Thank you so much!

JuniorCompressor · Accepted Answer · 2015-03-29 11:58:00Z

2

You can use filter to keep only the elements greater than n:

>>> def all_gt(nums, n):
...     return filter(lambda x: x > n, nums)
>>> all_gt([1, 2, 3, 4], 4)
[]
>>> all_gt([2, 3, 4, 5], 3)
[4, 5]

answered Mar 29, 2015 at 11:58

JuniorCompressor

20.1k4 gold badges33 silver badges58 bronze badges

4 Comments

Anupam Over a year ago

I love the use of functional programming part of python :) ,

JuniorCompressor Over a year ago

Everyone secretly loves it and hates it

icefish Over a year ago

Yes I know but my assignment needs me to do in .append :(

JuniorCompressor Over a year ago

Then Rao's answer should do

Anupam · Accepted Answer · 2015-03-29 12:29:10Z

0

just applying list comprehension with a if condition will do the trick

nums = [ 1, 2, 3, 4, 5]
constraint = 3

filtered_result = [ x for x in nums if x > constraint]

will result into [4,5]

answered Mar 29, 2015 at 12:29

Anupam

7371 gold badge14 silver badges28 bronze badges

Collectives™ on Stack Overflow

Constructing a list from another list in python

3 Answers 3

Your mistake

Better Ways

3 Comments

4 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Your mistake

Better Ways

3 Comments

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related