如何在 Python 的列表中的两个元素之间找到缺失的元素？

Question

I have a list as follows:

['1', '5', '6', '7', '10']

I want to find the missing element between two elements in the above list. For example, I want to get the missing elements between '1' and '5' , ie '2' , '3' and '4' . Another example, there are no elements between '5' and '6' , so it doesn't need to return anything.

Following the list above, I expect it to return a list like this:

['2', '3', '4', '8', '9']

My code:

# Sort elements in a list
input_list.sort()

# Remove duplicates from a list
input_list = list(dict.fromkeys(input_list))

How to return the above list? I would appreciate any help. Thank you in advance!

Answer 1

If you iterate in pairs, it is easy to detect and fill in the gaps:

L = ['1', '5', '6', '7', '10']
result = []
for left, right in zip(L, L[1:]):
    left, right = int(left), int(right)
    result += map(str, range(left + 1, right))

Answer 2

I would use the range() function to generate the missing numbers, and maybe use itertools.pairwise() to easily compare to the previous number. Since Python 3.10, pairwise is better than zip(arr, arr[1:]) because pairwise is implemented in the C layer, and does not make a copy of lists.

import itertools
arr = ['1', '5', '6', '7', '10']

new_arr = []

for p, c in itertools.pairwise(arr):
    prev, curr = int(p), int(c)
    if (prev + 1) != curr:
        new_arr.extend([str(i) for i in range(prev + 1, curr)])

Answer 3

If performance is not an issue, you can use a simple nested for loop:

input_list = ['1', '5', '6', '7', '10']

missing_list = []

for ind in range(0, len(input_list)-1):

    el_0 = int(input_list[ind])
    el_f = int(input_list[ind+1])

    for num in range(el_0 + 1, el_f):
        missing_list.append(str(num))

The above assumes that the numbers are integers. The first loop is something not recommended - the loop iterates using the length of the list instead of constructs like enumerate . I used it here for its simplicity.

Answer 4

You could use a combination of range and set and skip iteration altogether.

#original data
data = ['1', '5', '6', '7', '10']

#convert to list[int]
L    = list(map(int, data))

#get a from/to count
R    = range(min(L), max(L)+1)

#remove duplicates and convert back to list[str]
out  = list(map(str, set(R) ^ set(L)))

print(out)

Answer 5

Using a classic for loop and range() method and also for one-liner fans:

arr = ['1', '5', '6', '7', '10']
ans = []

for i in range(len(arr) - 1): ans += map(str, list(range(int(arr[i]) + 1, int(arr[i + 1]))))
    
print(ans)

outputs: ['2', '3', '4', '8', '9']