Multiple loop in list comprehension-CodePudding

I have a problem set in hand, the description is something like:

We have 2 lists ie "nums1" and "nums2" and 2 integers m and n

nums1 = [1, 2, 3, 7, 8, 9]
m = 3
nums2 = [2, 5, 6]
n = 2

Here "m" denotes number of elements to be picked from "nums1" and "n" denotes the number of elements to be picked from "nums2". There are other ways ofcourse, but I want to create a list with only "list comprehension" on the above conditions.

With a simple for loop I can achieve that with:

li = []
for i in range(m):
  li.append(nums1[i])
for j in range(n):
  li.append(nums2[j])
print(li)

I tried the following but the outcome result is not as expected:

[a for a in nums1 for b in nums2]

Expected output: [1, 2, 2, 3, 5]

Also, does list comprehension supports multiple loop?

CodePudding user response：

Given the following:

nums1 = [1, 2, 3, 7, 8, 9]
m = 3
nums2 = [2, 5, 6]
n = 2

Using strictly list comprehension, the following does what you asked for:

sorted([nums1[a] for a in range(m)]   [nums2[b] for b in range(n)])

Returns:

[1, 2, 2, 3, 5]

as per the conditions with "List comprehension" to be specific.

Although as others have said in the comment, a slice would have been more pythonic and easier to reason about. But since you asked for a list comprehension implementation, then this is as close to an answer as you'll get.

CodePudding user response：

nums1 = [1, 2, 3, 7, 8, 9]
m = 3
nums2 = [2, 5, 6]
n = 2

ll = sorted([a for b in [nums1[:m], nums2[:n]] for a in b])

print(ll)

Output

[1, 2, 2, 3, 5]

CodePudding user response：

nums1 = [1, 2, 3, 7, 8, 9]
m = 3
nums2 = [2, 5, 6]
n = 2

Create an empty list and store the returned values in this list if you need to access them later on in the program

nums3 = []
[nums3.append(nums1[i]) for i in range(m)]
[nums3.append(nums2[i]) for i in range(n)]

Sorts and prints the newly created list

print(sorted(nums3))