如何将csv文件中具有整数值的列拆分为新列

Question

I have a one column of numbers, like 22,23,25,32,33,36. and I want to split it new columns in csv file such that first column has 22, second column 23 and so on.

import csv
from csv import writer


COLUMNS = 6

with open("Winning No - Sheet1.csv", "r") as input:
    with open("output_file.csv", "w") as f:
        output = writer(f, delimiter=";")
        output.writerow(["Col {}".format(i+1) for i in xrange(COLUMNS)])
        buffer = []
        for row in input:
            buffer.append(row)
            if len(buffer) == COLUMNS:
                output.writerow(row.split(','))
                del buffer[:]

When I tried this, I get 6 columns with all the numbers in it such that first column with 22,23,25,32,33,36. Second column with another six numbers. How to sort this out?

Output that I want:

COL1 COL2 COL3 COL4 COL5 COL6 
22    23   25   32   33   36 

Output that I get is actually:

COL1        COL2          COL3       COL4     COL5       COL6 
22,23...,36 1,3,,...,36  3,4,...9  4,6,...7  8,9,...10  1,2,....7 

Answer 1

You probably want to do something like:

output.writerow(row.split(','))

I've never used the csv module, but it looks like it also provides a reader which you could use to do the parsing.

Updated with more context:

with open("Winning No - Sheet1.csv", "r") as input:
    with open("output_file.csv", "w") as f:
        output = writer(f, delimiter=";")
        output.writerow(["Col {}".format(i+1) for i in xrange(COLUMNS)])
        for row in input:
            output.writerow(row.split(','))