How to increment the strings in list with digits if the element already present in the list-CodePudding

I have a list of strings with repeated elements present. I have to recreate the list with a condition that if the string is already present in the list, I need to append a number to the end of the string.

Here is my list:

a = ['abc', 'abc', 'h', 'xv', 'xv', 'xv', 'h', 'h', 'h', 'h']

I expect the following output:

['abc', 'abc1', 'h', 'xv', 'xv1', 'xv2', 'h1', 'h2', 'h3', 'h4']

I tried to accomplish this with the following code:

new = []
for i in a:
    if i in new:
        if i[:-1].isdigit():
            new.append(i   str(int(i[-1])   1))
        else:
            new.append(i   '1')
    else:
        new.append(i)

But I'm getting:

['abc', 'abc1', 'h', 'xv', 'xv1', 'xv1', 'h1', 'h1', 'h1', 'h1']

But it is not giving the correct results. It will be great if I can get a list comprehension or an optimized one-liner for this problem.

CodePudding user response：

You can use a dictionary to keep track of the number of occurrences of a string seen so far. You could use a list comprehension, but the only way I could think of would involve list slicing and .count() or something super hacky, both of which would be less desirable than the one-pass implementation below:

a = ['abc', 'abc', 'h', 'xv', 'xv', 'xv', 'h', 'h', 'h', 'h']
result = []
occurrences = {}

for elem in a:
    if elem in occurrences:
        elem_to_add = elem   str(occurrences[elem])
        occurrences[elem]  = 1
    else:
        elem_to_add = elem
        occurrences[elem] = 1
    result.append(elem_to_add)

print(result)

CodePudding user response：

Use a dictionary to keep track of the items while you traverse a.

d = {}
out = []
for i in a:
    if i not in d:
        out.append(i)
        d[i] = 1
    else:
        out.append(i str(d[i]))
        d[i]  = 1

Output:

>> print(out)
['abc', 'abc1', 'h', 'xv', 'xv1', 'xv2', 'h1', 'h2', 'h3', 'h4']

CodePudding user response：

Please try with this

a = ['abc', 'abc', 'h', 'xv', 'xv', 'xv', 'h', 'h', 'h', 'h']

new = []
for i in a:
    if i in new:
        index = 1
        while i   str(index) in new:
            index  = 1
        if index == 1:
            new.append(i   '1')
        else:
            new.append(i   str(index))
            
    else:
        new.append(i)
        
print(new)

CodePudding user response：

if you dont want to create a dictionary to store and also don't want to create a new list but make edit in the original, you can use this:

count = 0
i = 0
for word in a:
    i  = 1
    for subindex in range(i,len(a)):
       if word == a[subindex]:
        count  = 1
        a[subindex] = word str(count)
    count = 0