Convert list to interger

[Clives]

Greetings,
The csv returns: [114436]
I want to get the values as an integer and not list, any pointers appreciated.
Thanks
Clive

##### CSV id = open(manhole, encoding='utf-8' ) manhole_id = csv.reader(id) next(id, None) rows = () for row in manhole_id: data_manhole_id = [int(item) for item in row] print(data_manhole_id)

I still get:

Output:
[114436] [114474] [114514]

Not

Output:
114436 114474 114514

I also get an error: TypeError
TypeError: list indices must be integers or slices, not list

[deanhystad]

I assume your file looks like this:
114436
114474
114514

This is not csv format. You do not need a CSV reader.

with open("data.csv", "r") as file: for line in file: print(int(line))

What you wrote is for data that is actually in csv format
114436,114474,114514

import csv with open('data.csv', "r") as file: reader = csv.reader(file) for row in reader: data = [int(item) for item in row] print(data)

Output:
[114436, 114474, 114514]

[Clives]

Hi Dean,
Thanks for the advice.

I changed the code:

 line = () with open(data, encoding='utf-8') as manhole_id: #csv_reader = reader(manhole_id) for line in manhole_id: next(manhole_id, None)

The next(manhole_id, None) is not working??
It is returning an extra line??

Thanks
Clive

[deanhystad]

If the file looks like this:
Whatever
114436
114474
114514

You can skip the first line in the file.

with open("data.csv", "r") as file: next(file) for line in file: print(int(line))

Output:
114436 114474 114514

When using a contact manager (with), the file only remains open for code that is in the context block (indented 1 level deeper than the with). In your latest example the manhole_id file is closed before the for loop executes.

[Clives]

Apologies typo...

It returns:
114806

114810

114831

114841

114843

NOT:
114806
114810
114831
114841
114843

[deanhystad]

You are doing something wrong. My guess is you are printing the str which has a trailing linefeed. I thought you wanted to read the file and convert to int.