How to iterate over every n-th line from a file?

Question

First, I am new to Python and did search for answers, but no luck. So far what I found only returns a one line like my code below. I tried other solutions like itertools.islice but always only get a one line back.

I have a file called data.txt containing lines of data:

This is line one
This is line two 
This is line three 
This is line four 
This is line five 
This is line six 
This is line seven 
...

I have the following code:

with open('data.txt', 'r') as f:
    for x, line in enumerate(f):
        if x == 3:
            print(line)

In this case it only prints

"This is line four".

I do understand why but how do I take it from here and have it print the lines 4, 7, 10, 13, ...?

Answer 1

The return value of open is an iterator (and thus, iterable), so you can pass it to itertools.islice :

islice(iterable, start, stop[, step]) --> islice object
Return an iterator whose next() method returns selected values from an iterable. [...]

Demo:

data.txt :

line1
line2
line3
line4
line5
line6
line7
line8
line9
line10
line11
line12
line13

Code:

from itertools import islice

with open('data.txt') as f:
    for line in islice(f, 3, None, 3):
        print line,  # Python3: print(line, end='')

Produces:

line4
line7
line10
line13

Answer 2

In Your code You are printing when x == 3 , so You are printing only the fourth line in a file, because enumeration starts with 0 .

Try:

with open('data.txt', 'r') as f:
    for x, line in enumerate(f):
        if x%3 == 1:
            print(line)

x%3 == 1 means that rest from division x by 3 has to be 1 . This way You will print the lines 1, 4, 7, etc. .

Answer 3

You need to use https://en.wikipedia.org/wiki/Modular_arithmetic

with open('data.txt', 'r') as f:
    for x, line in enumerate(f):
        if x and x % 3 == 0:
            print(line)