Python: Never entering if statement for conversion [duplicate]

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• Fastest way to check if a value exist in a list 7 answers

I have an list which tells me which columns need to be converted from bytes into KB.

covert_to_kb_columns=[9, 10, 21, 22, 24]

I am having trouble implementing my code correctly. I want to check that if I am on the an 'idx' number that matches one of the numbers in the covert_to_kb_columns list than I should preform the calculation otherwise leave the number as is. The try/except statement is for when there is no number in the file just a '', so I append a zero.

print covert_to_kb_columns
for idx, column_number in enumerate(columns_I_am_working_with):
    print idx
    #Check if I need to convert number from bytes to KB
    if idx == [number for number in covert_to_kb_columns]:
        print "Need to divide by 1024.0"
        data_to_use.append("{:.1f}".format(float(row[column_number]) / 1024.0))
    #Otherwise just append number as is
    else:
        try:
            float(row[column_number])
            data_to_use.append(row[column_number])
        except ValueError:
            data_to_use.append('0')

My print statements give me this result: (Note '.' is to indicate that all numbers were printed but I'm not writing them all 39 numbers out )

[9, 10, 21, 22, 24]
0
1
2
3
.
.
.
9
10
.
.
.
21
22
23
24
.
.
.
39

It seems that it never enters the print statement to perform the conversion. I think the issue is with the line

if idx == [number for number in covert_to_kb_columns]:

But I can't spot what I doing wrong