Replacing values in csv by using python

Question

def update(all_marks, stud_num, mark, column, result):
    lines = [l for l in all_marks]

    for row in all_marks:
        if stud_num in row:
            lines[rows][column] = mark

From here, I am trying to replace the value by using lines[rows][column] = mark.

It is supposed to replace the existing value with mark. But there's a problem with defining rows. Anyone knows how to fix? Thanks.

Edit: Here's sample of data from all_marks:

[['a', '', '', '', '', ''],

['b', '', '', '', '', ''],

['c', '', '', '', '', ''],

['d', '', '', '', '', ''],

['e', '', '', '', '', ''],

['f', '', '', '', '', ''],

['g', '', '', '', '', '']]

What I want to do here is to replace the value in '' with mark.

For example, def update(all_marks, 'a', '10', 2, True): will return

[['a', '', '10', '', '', ''],

['b', '', '', '', '', ''],

['c', '', '', '', '', ''],

['d', '', '', '', '', ''],

['e', '', '', '', '', ''],

['f', '', '', '', '', ''],

['g', '', '', '', '', '']]

Thanks for helping a newbie.

Please take care to format your code properly. The one that you posted won't even run. Also what exact problem do you encounter? — Ihor Kaharlichenko
– Ihor Kaharlichenko, Commented Aug 19, 2012 at 7:29
Did you try the Python csv modul? It works fine for tasks like yours. — MaxPowers
– MaxPowers, Commented Aug 19, 2012 at 8:07

Dreen · Accepted Answer · 2012-08-19 10:29:48Z

2

Here is a modified version of your function that will return the output as expected:

def update(all_marks, stud_num, mark, column):
    for i in range(len(all_marks)):
        if stud_num in all_marks[i]:
            all_marks[i][column] = mark
    return all_marks

And here is how it works:

>>> marks
[['a', '', '', '', '', ''], ['b', '', '', '', '', ''], ['c', '', '', '', '', ''], ['d', '', '', '', '', ''], ['e', '', '', '', '', ''], ['f', '', '', '', '', ''], ['g', '', '', '', '', '']]

>>> update(marks,'a','10',2)
[['a', '', '10', '', '', ''], ['b', '', '', '', '', ''], ['c', '', '', '', '', ''], ['d', '', '', '', '', ''], ['e', '', '', '', '', ''], ['f', '', '', '', '', ''], ['g', '', '', '', '', '']]

Note that marks is now modified

>>> marks
[['a', '', '10', '', '', ''], ['b', '', '', '', '', ''], ['c', '', '', '', '', ''], ['d', '', '', '', '', ''], ['e', '', '', '', '', ''], ['f', '', '', '', '', ''], ['g', '', '', '', '', '']]

If you want to change that so that update simply returns a copy of modified data change the function in the following way:

def update(all_marks, stud_num, mark, column):
    tmp = all_marks
    for i in range(len(tmp)):
        if stud_num in tmp[i]:
            tmp[i][column] = mark
    return tmp

edited Aug 19, 2012 at 10:29

answered Aug 19, 2012 at 10:20

Dreen

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1 Comment

Dreen Over a year ago

I initially posted the code and just updated my answer to give you some more information

Jealie · Accepted Answer · 2012-08-19 10:21:50Z

1

Here is the working code:

def update(all_marks, stud_num, mark, column, result):
    lines = [l for l in all_marks]
    for row in range(len(all_marks)):
        if all_marks[row][0] == stud_num:
            lines[row][column] = mark

And here are the explanations:

for row in range(len(all_marks)):

=> you don't want to iterate over list objects (e.g. ['a','','','','','']) but over list indices

if stud_num == all_marks[row][0]:

=> This is to check only the first character of your row, not any character.

lines[row][column] = mark

=> typo here, should be row and not rows

answered Aug 19, 2012 at 10:21

Jealie

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1 Comment

Jealie Over a year ago

Dreen beat me up for a few seconds on this one :)

Pierre GM · Accepted Answer · 2012-08-19 10:54:31Z

0

Looping on the row indices isn't as efficient as looping on the rows themselves. A more pythonic way would be

def update2(all_marks,stud_num,mark,column):
    for row in all_marks:
        if stud_num in row:
            row[column] = mark
    return all_marks

answered Aug 19, 2012 at 10:54

Pierre GM

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