Question

I need to remove a line from a csv with a specific pattern

this is how my csv file looks.

lbm,16730,0
namd,16733,6
namd,16731,2
namd,16732,4

If I want to remove the line with the pattern 16730 and output the rest of the file as it is..

so, the output something like this:

namd,16733,6
namd,16731,2
namd,16732,4

how do I do that?

here is a small script I wrote with the help of some files on the internet

def delete_line(dello):
    opener = open(input_csv, 'rb')
    dataset = csv.reader(opener, delimiter=',')
    output = []
    for line in dataset:
            if 'dello' == line[1]:
                    print line[1]
                    #i dont know how to strip it here
                    output.append(line)
    opener.close()
    fn = input_csv
    f = open(fn,'w')
    f.writelines(output)
    f.close()

any hints where I am going wrong?

Answer 1

If you need python, then use this:

def delete_line(dello):
    data = open("abc.csv").readlines()

    i = 0
    for line in data:
        if dello in line:
            data.pop(i)
        i += 1

    open("abc.csv", "w").write("".join(data))

delete_line("16732")

Input:

lbm,16730,0
namd,16733,6
namd,16731,2
namd,16732,4

Output:

lbm,16730,0
namd,16733,6
namd,16731,2

Note: this will remove all the entries matching the string.

---

Update

Modifying your code:

import csv

def delete_line(dello):
    opener = open("abc.csv", 'rb')
    dataset = csv.reader(opener, delimiter=',')
    output = []
    for line in dataset:
        # Add to output only if not matching the string
        if dello != line[1]:
            # Need join as line is a list
            output.append(",".join(line) + "\n")
    opener.close()

    fn = "abc.csv"
    f = open(fn,'w')
    f.writelines(output)
    f.close()

delete_line("16730")

If you need to strip out an entry, you can use dataset.pop(index).