How can i get all the lines after the first in a file python?

Question

i would like to know how i could get all lines after the first in a python file

I've tried with this:

fr = open("numeri.txt", "r")

count = 0
while True:
    line = fr.readline(count)
    if line == "":
        break
    count += 1
    print(line)
fr.close()

Could anyone help me? Thanks

Does this answer your question? [Read file from line 2 or skip header row](https://stackoverflow.com/questions/4796764/read-file-from-line-2-or-skip-header-row) — Tomerikoo, Dec 03 '20 at 15:17

score 0 · Answer 1 · answered Dec 03 '20 at 15:04

0

You could add an extra if statement to check if count != 0 Since on the first loop it will be 0.

answered Dec 03 '20 at 15:04

xl3ehindTim

121
1
7

score 0 · Answer 2 · answered Dec 03 '20 at 15:08

0

I don't know if i understood well, but to obtain all the lines skipping the first one you can simple do

lines = []
with open("numeri.txt") as fobj:
   lines = fobj.readlines()[1:] 
count = len(lines)+1 if lines else 0 # If you want to maintain the same counting as in your example

answered Dec 03 '20 at 15:08

Cristian

119
3

The question is to iterate the lines after the first, not count them... – Tomerikoo Dec 03 '20 at 15:21
it doesn't say iterate all lines, it says get all lines after the first one – Cristian Dec 04 '20 at 07:41

Riccardo Bucco · Answer 3 · 2020-12-03T15:16:00.493

0

Just read the first line without using it:

with open('numeri.txt') as f:
    f.readline()
    lines = f.readlines()
print(*lines, sep='')

To ignore the first line you can also use next(f) (instead of f.readline()).

This is also fine:

with open('numeri.txt') as f:
    lines = f.readlines()[1:]
print(*lines, sep='')

edited Dec 03 '20 at 15:16

answered Dec 03 '20 at 15:10

Riccardo Bucco

13,980
4
22
50

score 0 · Answer 4 · answered Dec 03 '20 at 15:10

count = 0
with open(file, 'r') as file:
    next(file.readline())  # skip the first line
    for count, line in enumerate(file.readlines()): # read remaining lines with count
        if not line:  # If line equals ""  this will be True
            break
        print(count, line)
  
count -= 1 # To ignore last lines count.

Keredu · Answer 5 · 2020-12-03T15:24:12.490

0

Try using l[1:]. It returns a subset of l that consist in the elements of l except the first position.

with open("numeri.txt", "r") as f:
    content = f.readlines()[1:]
for line in content:
    print(line.strip('\n')) # In order to avoid introduce double \n since print ends with a '\n'

EDIT: Based on @riccardo-bucco ' solution:

with open("numeri.txt", "r") as f:
    content = f.readlines()[1:]
    print(*content, sep='')

edited Dec 03 '20 at 15:24

answered Dec 03 '20 at 15:12

Keredu

378
1
14

`print(*content)` is enough, and won't add double `\n` characters ;) – Riccardo Bucco Dec 03 '20 at 15:13
1

It introduces a whitespace after every line. – Keredu Dec 03 '20 at 15:15
2

Then use `print(*content, sep='')` – Riccardo Bucco Dec 03 '20 at 15:17
I made it explicitly for better understanding of the comment on the right side. – Keredu Dec 03 '20 at 15:25

H S · Answer 6 · 2020-12-03T16:31:40.130

0

To print all but the first line:

with open('numeri.txt', 'r') as f:
    output = ''.join(f.readlines()[1:])
    print(output)

edited Dec 03 '20 at 16:31

answered Dec 03 '20 at 15:24

H S

45
4

score -1 · Answer 7 · answered Dec 03 '20 at 15:06

-1

start count at 1 so it skips the first line

...
count = 1
...

answered Dec 03 '20 at 15:06

Hadrian

917
5
10

How can i get all the lines after the first in a file python?

7 Answers7