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.

• Stefan van den Akker: 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.
• YXD: 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.

Answer 1

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]

• icefish: The return num2 return nothing.
• Bhargav Rao: @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: Oh my mistake, I still put a return infront of num2.append(i). It works now. Thank you so much!

Answer 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]

• Anupam: I love the use of functional programming part of python :) ,
• JuniorCompressor: Everyone secretly loves it and hates it
• icefish: Yes I know but my assignment needs me to do in .append :(
• JuniorCompressor: Then Rao's answer should do

Answer 3

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]