Nested For Loop Counter

Question

I am very beginner with Python and am having trouble with a certain aspect of the counter as it relates to its use in a nested for loop.

I am trying to run a nested for loop that checks if an array A has any duplicate values.

Trying to talk myself (and y'all) through this to make sense out of it: I am using a nested for loop to essentially loop through each item in array A...and for each item in array A, I need another counter to loop A so that I can compare A to itself in the form of counter i and counter j. Here is the issue: I don't want to count on itself aka i don't want to double count. And if I simply type the code that y'all will see below, it will double count (count on itself). So I want to make sure that the index of my inner for loop's counter is always +1 to my outer for loops counter.

Here is what the code looks like:

A = [4, 3, 2, 4]

for i in A:
    for j in A:
        if i == j:
            print("yup")

the output is...you guessed it:

yup
yup
yup
yup
yup
yup

6 "yup"'s because each time it is counting each number on itself.

Hopefully I am explaining that properly...

So my question is: does anybody know how to make sure that my "j" counter is indexed +1...

I thought it would be:

for i in A:
    for j = i + 1 in A:
        if i == j:
            print("yup")

but apparently that isn't right

Any insight here is very much appreciated!!!

Thanks, Mark

use enumerate() with for loops for indexing. If still confused go with while loops and do natural indexing — JenilDave
– JenilDave, Commented May 15, 2020 at 3:41

Shubham Gupta · Accepted Answer · 2020-05-15 03:29:39Z

2

If you want to start j from 1 you can simply use the range function

for i in range(len(A)):
    for j in range(i+1,len(A)):
        if A[i] == A[j]:
            print("yup")

Hope this is what you are looking for.

answered May 15, 2020 at 3:29

Shubham Gupta

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1 Comment

Shubham Gupta Over a year ago

If you are counting duplicates in a list, I would suggest you use Counter or HashMap which will reduce your time complexity to O(n). You can check out this link

João Castilho · Accepted Answer · 2020-05-15 03:32:12Z

1

This is a solution for your problem. enumerate function returns a tuple with the index and value of each array element.

for idx_i, elem_i in enumerate(A):
  for idx_j, elem_j in enumerate(A[idx_i+1:]):
      if elem_i == elem_j:
         print("Yup")

If you want an array with unique elements, there are better efficient ways to do that

answered May 15, 2020 at 3:32

João Castilho

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Mark · Accepted Answer · 2020-05-15 03:34:15Z

1

You can use enumerate to get the current index in the outside loop and then loop over a slice on the inside:

A = [4, 3, 2, 4]

for i, n_outer in enumerate(A):
    for n_inner in A[i+1:]:
        if n_inner == n_outer:
            print("yup")

edited May 15, 2020 at 3:34

answered May 15, 2020 at 3:31

Mark

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