Incorporating instance 'count' into list joined into string (Python3)

Question

First time asking a question on SO, so please bear with me.

In the python program I'm writing, I am trying to avoid concatenation of a string in a for loop by using the join method (because: efficiency), but I would like for the program to put a count next to each instance.

For example:

[1] First item [2] Second item [3] Third item

From what I understand (I'm not a master pythoner), this form of using .join() is utilizing a list comprehension, and I can't seem to figure out the proper syntax to do what I'm trying to.

temporary_list = ["First item", "Second item", "Third item"]
count = 1
display = "\n".join("> [{}] {}".format((count), t), count += 1 for t in temporary_list[:10])

I could easily do this with a for loop:

    for t in temporary_list[:10]:
            display += "> [{}] {}\n".format(count, t)
            count += 1

I have a similar line of code that works fine without counting.

display = "\n".join("> {}: {}".format(t, display_correctly(number_list[t])) for t in temporary_list[:10])

I've tried playing with the parenthesis, using 'and' and a lot of googling. Is there a way to do this?

If it matters, I'm using python 3.6

Chris · Accepted Answer · 2020-03-03 02:03:04Z

5

IIUC, you need enumerate:

temporary_list = ["First item", "Second item", "Third item"]
print("\n".join("> [{}] {}".format(n, i) for n, i in enumerate(temporary_list, start=1)))

Output:

> [1] First item
> [2] Second item
> [3] Third item

answered Mar 3, 2020 at 2:03

Chris

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1 Comment

downwithocp Over a year ago

This works perfectly. Enumerate was the missing link. Thank you.

lutre69 · Accepted Answer · 2020-03-03 02:11:10Z

0

Try:

"\n".join("> [{}] {}".format((count), t) for count, t in zip(range(len(temporary_list[:10])), temporary_list[:10]))

My bad, this works now

edited Mar 3, 2020 at 2:11

answered Mar 3, 2020 at 1:59

lutre69

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2 Comments

downwithocp Over a year ago

That was fast. It was missing a couple of parenthesis, but I ran it and got: TypeError: 'list' object cannot be interpreted as an integer

lutre69 Over a year ago

Sorry I fixed the code, but enumerate looks way better I admit.

Errol · Accepted Answer · 2020-03-03 02:14:12Z

0

If you just want to print out the index of items in the list. But if you want to count the number of occurences of each item in a list use Counter()

temporary_list = ["First item", "Second item", "Third item"]

for index,value in enumerate(temporary_list):
    print(f'[{index+1}]', value)

OUTPUT:

[1] First item
[2] Second item
[3] Third item

If you want to print out horizontally:

temporary_list = ["First item", "Second item", "Third item"]
print(" ".join([f'[{index+1}] {value}' for index,value in enumerate(temporary_list)]))

OUTPUT:

[1] First item [2] Second item [3] Third item

edited Mar 3, 2020 at 2:14

answered Mar 3, 2020 at 2:05

Errol

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3 Comments

downwithocp Over a year ago

The counting was strictly for displaying purposes and not stored. Your code answers the question, though, thank you.

Errol Over a year ago

Sorry, your title say occurrence count. I was thinking you want to count the occurrences of each items in the list.

downwithocp Over a year ago

I apologize for my limitations in figuring out a clearer wording. The method this code is built into has a user-defined list. If that list has more than 10 items, it's broken into different pages. I thought it would look nicer to have an index alongside each entry (and if it goes to the second page, the first item would be 1 again instead of 11). That's what I meant by counting.

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