Getting slightly different output every time after running the same code

Question

Here in this code I am getting output differently every time I run the same code.

INPUT:

s='AABCAAADA'
st=[]
def merge_the_tools(size,k):
    n=int(len(size)/k)
    for i in range(n):
        st.append(size[i*n:(i+1)*n])
    for i in st:
        se=set(i)
        print(''.join(se))
        
print(merge_the_tools(s,3))

First OUTPUT:

AB
AC
AD
None

Another OUTPUT:

AB
CA
DA
None

Another OUTPUT:

BA
CA
DA
None

Like this I am getting different output Can anybody tell why this is happening.

And I want this OUTPUT:

AB
CA
AD

Sort the content of the set to obtain always the same order. — dlask
– dlask, Commented Oct 16, 2021 at 6:49

Gaberocksall · Accepted Answer · 2021-10-16 06:49:56Z

1

Python sets have no order, so there is no guarantee that the items will be retrieved in the same order every time while joining them. Consider using a list or tuple if you would like to maintain order.

answered Oct 16, 2021 at 6:49

Gaberocksall

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Troll · Accepted Answer · 2021-10-16 06:56:11Z

1

Sets in Python are unordered and unindexed as shown in this demonstration. If you want to maintain a specific order, you can just sort the set as follows.

print(''.join(sorted(se)))

answered Oct 16, 2021 at 6:56

Troll

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justANewb stands with Ukraine · Accepted Answer · 2021-10-16 07:22:30Z

0

A set in Python is an unordered data structure, so so it does not preserve the insertion order.

You should use a list instead. Or you can sort the set. But sort return a list anyway.

Using sorted:

s='AABCAAADA'
st=[]
def merge_the_tools(size,k):
    n=int(len(size)/k)
    for i in range(n):
        st.append(size[i*n:(i+1)*n])
    for i in st:
        se=set(i)
        print(''.join(sorted(se)))
        
print(merge_the_tools(s,3))

edited Oct 16, 2021 at 7:22

answered Oct 16, 2021 at 6:51

justANewb stands with Ukraine

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