How to print every nth index of a python list on a new line?

I'm trying to print out a list and for every 5 indexes, it prints a new line. So for example, if I have:

  [1,2,3,4,5,6,7,8,9,10]

the output would be:

  1 2 3 4 5

  6 7 8 9 10

I tried this so far:

   lst = [1,2,3,4,5,6,7,8,9,10]



   for i in lst:

       if len(lst) > 5:

          print(lst,'n')

but all I get is:

   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   .......

How could I do this?

Thanks for the help!

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

asked Jan 22 at 5:06

Landon G

12212

migrated from superuser.com Jan 22 at 5:25

This question came from our site for computer enthusiasts and power users.

What's your output if len(list) isn't divisible by your index?

– Primusa
Jan 22 at 5:46

add a comment |

I'm trying to print out a list and for every 5 indexes, it prints a new line. So for example, if I have:

  [1,2,3,4,5,6,7,8,9,10]

the output would be:

  1 2 3 4 5

  6 7 8 9 10

I tried this so far:

   lst = [1,2,3,4,5,6,7,8,9,10]



   for i in lst:

       if len(lst) > 5:

          print(lst,'n')

but all I get is:

   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   .......

How could I do this?

Thanks for the help!

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

asked Jan 22 at 5:06

Landon G

12212

migrated from superuser.com Jan 22 at 5:25

This question came from our site for computer enthusiasts and power users.

What's your output if len(list) isn't divisible by your index?

– Primusa
Jan 22 at 5:46

add a comment |

I'm trying to print out a list and for every 5 indexes, it prints a new line. So for example, if I have:

  [1,2,3,4,5,6,7,8,9,10]

the output would be:

  1 2 3 4 5

  6 7 8 9 10

I tried this so far:

   lst = [1,2,3,4,5,6,7,8,9,10]



   for i in lst:

       if len(lst) > 5:

          print(lst,'n')

but all I get is:

   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   .......

How could I do this?

Thanks for the help!

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

asked Jan 22 at 5:06

Landon G

12212

I'm trying to print out a list and for every 5 indexes, it prints a new line. So for example, if I have:

  [1,2,3,4,5,6,7,8,9,10]

the output would be:

  1 2 3 4 5

  6 7 8 9 10

I tried this so far:

   lst = [1,2,3,4,5,6,7,8,9,10]



   for i in lst:

       if len(lst) > 5:

          print(lst,'n')

but all I get is:

   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 



   .......

How could I do this?

Thanks for the help!

python list printing split

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

asked Jan 22 at 5:06

Landon G

12212

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

asked Jan 22 at 5:06

Landon G

12212

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

edited Jan 22 at 5:48

Mehrdad Pedramfar

6,34411543

asked Jan 22 at 5:06

Landon G

12212

asked Jan 22 at 5:06

Landon G

12212

asked Jan 22 at 5:06

Landon G

12212

migrated from superuser.com Jan 22 at 5:25

This question came from our site for computer enthusiasts and power users.

migrated from superuser.com Jan 22 at 5:25

This question came from our site for computer enthusiasts and power users.

What's your output if len(list) isn't divisible by your index?

– Primusa
Jan 22 at 5:46

add a comment |

What's your output if len(list) isn't divisible by your index?

– Primusa
Jan 22 at 5:46

What's your output if len(list) isn't divisible by your index?

– Primusa
Jan 22 at 5:46

add a comment |

4 Answers
4

active

oldest

votes

Try this:

a = [1,2,3,4,5,6,7,8,9,10]

for i in [a[c:c+5] for c in range(0,len(a),5) if c%5 == 0]:

    print(*i)

the output will be:

1 2 3 4 5

6 7 8 9 10

also you can replace 5 with any other number or variables.

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

Hi, do you mind if I use the *i argument unpacking for the formatting in my answer? It's a lot better than the current " ".join() i've got going on

– Primusa
Jan 22 at 5:35

@Primusa Sure it is ok, I will also vote-up your answer.

– Mehrdad Pedramfar
Jan 22 at 5:36

No problem @Primusa, I think that is good if you look at sep= in print function too. it is good. something like this print(*i, sep='-') it will print: 1-2-3-4-5

– Mehrdad Pedramfar
Jan 22 at 5:44

add a comment |

Used a for loop with a step:

n_indices = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for i in range(0, len(lst), n_indices):

    print(lst[i:i+n_indices])



>>>[1, 2, 3, 4, 5]

>>>[6, 7, 8, 9, 10]

If you want to be fancier with the formatting you can use argument unpacking like so: print(*list[i:i+n_indices]) and get outputs in this format:

1 2 3 4 5

6 7 8 9 10

edited Jan 22 at 5:37

answered Jan 22 at 5:31

Primusa

8,0002932

I guess he wants a formatted output only .join: print(' '.join(lst[i:i+indices]) ... will be good for him :)

– Puneet Sinha
Jan 22 at 5:43

add a comment |

You Can use below code :-

for i in range(0,len(a),5):

    print(" ".join(map(str, a[i:i+5])))

It will give you desired output.

Code Snip

edited Jan 22 at 5:46

Primusa

8,0002932

answered Jan 22 at 5:44

rachit17

242

add a comment |

A bit more idiomatic than the other answers:

n = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for group in zip(*[iter(lst)] * n):

    print(*group)

1 2 3 4 5

6 7 8 9 10

For larger lists, it's also much faster:

In [1]: lst = range(1, 10001)



In [2]: n = 5



In [3]: %%timeit

    ...: for group in zip(*[iter(lst)] * n):

    ...:     group

    ...:

236 µs ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)



In [4]: %%timeit

    ...: for i in range(0, len(lst), n):

    ...:     lst[i:i+n]

    ...:

1.32 ms ± 184 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

edited Jan 22 at 5:48

answered Jan 22 at 5:39

Tomothy32

7,8461728

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Try this:

a = [1,2,3,4,5,6,7,8,9,10]

for i in [a[c:c+5] for c in range(0,len(a),5) if c%5 == 0]:

    print(*i)

the output will be:

1 2 3 4 5

6 7 8 9 10

also you can replace 5 with any other number or variables.

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

Hi, do you mind if I use the *i argument unpacking for the formatting in my answer? It's a lot better than the current " ".join() i've got going on

– Primusa
Jan 22 at 5:35

@Primusa Sure it is ok, I will also vote-up your answer.

– Mehrdad Pedramfar
Jan 22 at 5:36

No problem @Primusa, I think that is good if you look at sep= in print function too. it is good. something like this print(*i, sep='-') it will print: 1-2-3-4-5

– Mehrdad Pedramfar
Jan 22 at 5:44

add a comment |

Try this:

a = [1,2,3,4,5,6,7,8,9,10]

for i in [a[c:c+5] for c in range(0,len(a),5) if c%5 == 0]:

    print(*i)

the output will be:

1 2 3 4 5

6 7 8 9 10

also you can replace 5 with any other number or variables.

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

Hi, do you mind if I use the *i argument unpacking for the formatting in my answer? It's a lot better than the current " ".join() i've got going on

– Primusa
Jan 22 at 5:35

@Primusa Sure it is ok, I will also vote-up your answer.

– Mehrdad Pedramfar
Jan 22 at 5:36

No problem @Primusa, I think that is good if you look at sep= in print function too. it is good. something like this print(*i, sep='-') it will print: 1-2-3-4-5

– Mehrdad Pedramfar
Jan 22 at 5:44

add a comment |

Try this:

a = [1,2,3,4,5,6,7,8,9,10]

for i in [a[c:c+5] for c in range(0,len(a),5) if c%5 == 0]:

    print(*i)

the output will be:

1 2 3 4 5

6 7 8 9 10

also you can replace 5 with any other number or variables.

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

Try this:

a = [1,2,3,4,5,6,7,8,9,10]

for i in [a[c:c+5] for c in range(0,len(a),5) if c%5 == 0]:

    print(*i)

the output will be:

1 2 3 4 5

6 7 8 9 10

also you can replace 5 with any other number or variables.

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

answered Jan 22 at 5:32

Mehrdad Pedramfar

6,34411543

Hi, do you mind if I use the *i argument unpacking for the formatting in my answer? It's a lot better than the current " ".join() i've got going on

– Primusa
Jan 22 at 5:35

@Primusa Sure it is ok, I will also vote-up your answer.

– Mehrdad Pedramfar
Jan 22 at 5:36

No problem @Primusa, I think that is good if you look at sep= in print function too. it is good. something like this print(*i, sep='-') it will print: 1-2-3-4-5

– Mehrdad Pedramfar
Jan 22 at 5:44

add a comment |

Hi, do you mind if I use the *i argument unpacking for the formatting in my answer? It's a lot better than the current " ".join() i've got going on

– Primusa
Jan 22 at 5:35

@Primusa Sure it is ok, I will also vote-up your answer.

– Mehrdad Pedramfar
Jan 22 at 5:36

No problem @Primusa, I think that is good if you look at sep= in print function too. it is good. something like this print(*i, sep='-') it will print: 1-2-3-4-5

– Mehrdad Pedramfar
Jan 22 at 5:44

Hi, do you mind if I use the *i argument unpacking for the formatting in my answer? It's a lot better than the current " ".join() i've got going on

– Primusa
Jan 22 at 5:35

@Primusa Sure it is ok, I will also vote-up your answer.

– Mehrdad Pedramfar
Jan 22 at 5:36

No problem @Primusa, I think that is good if you look at sep= in print function too. it is good. something like this print(*i, sep='-') it will print: 1-2-3-4-5

– Mehrdad Pedramfar
Jan 22 at 5:44

add a comment |

Used a for loop with a step:

n_indices = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for i in range(0, len(lst), n_indices):

    print(lst[i:i+n_indices])



>>>[1, 2, 3, 4, 5]

>>>[6, 7, 8, 9, 10]

If you want to be fancier with the formatting you can use argument unpacking like so: print(*list[i:i+n_indices]) and get outputs in this format:

1 2 3 4 5

6 7 8 9 10

edited Jan 22 at 5:37

answered Jan 22 at 5:31

Primusa

8,0002932

I guess he wants a formatted output only .join: print(' '.join(lst[i:i+indices]) ... will be good for him :)

– Puneet Sinha
Jan 22 at 5:43

add a comment |

Used a for loop with a step:

n_indices = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for i in range(0, len(lst), n_indices):

    print(lst[i:i+n_indices])



>>>[1, 2, 3, 4, 5]

>>>[6, 7, 8, 9, 10]

If you want to be fancier with the formatting you can use argument unpacking like so: print(*list[i:i+n_indices]) and get outputs in this format:

1 2 3 4 5

6 7 8 9 10

edited Jan 22 at 5:37

answered Jan 22 at 5:31

Primusa

8,0002932

I guess he wants a formatted output only .join: print(' '.join(lst[i:i+indices]) ... will be good for him :)

– Puneet Sinha
Jan 22 at 5:43

add a comment |

Used a for loop with a step:

n_indices = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for i in range(0, len(lst), n_indices):

    print(lst[i:i+n_indices])



>>>[1, 2, 3, 4, 5]

>>>[6, 7, 8, 9, 10]

If you want to be fancier with the formatting you can use argument unpacking like so: print(*list[i:i+n_indices]) and get outputs in this format:

1 2 3 4 5

6 7 8 9 10

edited Jan 22 at 5:37

answered Jan 22 at 5:31

Primusa

8,0002932

Used a for loop with a step:

n_indices = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for i in range(0, len(lst), n_indices):

    print(lst[i:i+n_indices])



>>>[1, 2, 3, 4, 5]

>>>[6, 7, 8, 9, 10]

If you want to be fancier with the formatting you can use argument unpacking like so: print(*list[i:i+n_indices]) and get outputs in this format:

1 2 3 4 5

6 7 8 9 10

edited Jan 22 at 5:37

answered Jan 22 at 5:31

Primusa

8,0002932

edited Jan 22 at 5:37

answered Jan 22 at 5:31

Primusa

8,0002932

answered Jan 22 at 5:31

Primusa

8,0002932

answered Jan 22 at 5:31

Primusa

8,0002932

I guess he wants a formatted output only .join: print(' '.join(lst[i:i+indices]) ... will be good for him :)

– Puneet Sinha
Jan 22 at 5:43

add a comment |

I guess he wants a formatted output only .join: print(' '.join(lst[i:i+indices]) ... will be good for him :)

– Puneet Sinha
Jan 22 at 5:43

I guess he wants a formatted output only .join: print(' '.join(lst[i:i+indices]) ... will be good for him :)

– Puneet Sinha
Jan 22 at 5:43

add a comment |

You Can use below code :-

for i in range(0,len(a),5):

    print(" ".join(map(str, a[i:i+5])))

It will give you desired output.

Code Snip

edited Jan 22 at 5:46

Primusa

8,0002932

answered Jan 22 at 5:44

rachit17

242

add a comment |

You Can use below code :-

for i in range(0,len(a),5):

    print(" ".join(map(str, a[i:i+5])))

It will give you desired output.

Code Snip

edited Jan 22 at 5:46

Primusa

8,0002932

answered Jan 22 at 5:44

rachit17

242

add a comment |

You Can use below code :-

for i in range(0,len(a),5):

    print(" ".join(map(str, a[i:i+5])))

It will give you desired output.

Code Snip

edited Jan 22 at 5:46

Primusa

8,0002932

answered Jan 22 at 5:44

rachit17

242

You Can use below code :-

for i in range(0,len(a),5):

    print(" ".join(map(str, a[i:i+5])))

It will give you desired output.

Code Snip

edited Jan 22 at 5:46

Primusa

8,0002932

answered Jan 22 at 5:44

rachit17

242

edited Jan 22 at 5:46

Primusa

8,0002932

edited Jan 22 at 5:46

Primusa

8,0002932

edited Jan 22 at 5:46

Primusa

8,0002932

answered Jan 22 at 5:44

rachit17

242

answered Jan 22 at 5:44

rachit17

242

answered Jan 22 at 5:44

rachit17

242

add a comment |

A bit more idiomatic than the other answers:

n = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for group in zip(*[iter(lst)] * n):

    print(*group)

1 2 3 4 5

6 7 8 9 10

For larger lists, it's also much faster:

In [1]: lst = range(1, 10001)



In [2]: n = 5



In [3]: %%timeit

    ...: for group in zip(*[iter(lst)] * n):

    ...:     group

    ...:

236 µs ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)



In [4]: %%timeit

    ...: for i in range(0, len(lst), n):

    ...:     lst[i:i+n]

    ...:

1.32 ms ± 184 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

edited Jan 22 at 5:48

answered Jan 22 at 5:39

Tomothy32

7,8461728

add a comment |

A bit more idiomatic than the other answers:

n = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for group in zip(*[iter(lst)] * n):

    print(*group)

1 2 3 4 5

6 7 8 9 10

For larger lists, it's also much faster:

In [1]: lst = range(1, 10001)



In [2]: n = 5



In [3]: %%timeit

    ...: for group in zip(*[iter(lst)] * n):

    ...:     group

    ...:

236 µs ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)



In [4]: %%timeit

    ...: for i in range(0, len(lst), n):

    ...:     lst[i:i+n]

    ...:

1.32 ms ± 184 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

edited Jan 22 at 5:48

answered Jan 22 at 5:39

Tomothy32

7,8461728

add a comment |

A bit more idiomatic than the other answers:

n = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for group in zip(*[iter(lst)] * n):

    print(*group)

1 2 3 4 5

6 7 8 9 10

For larger lists, it's also much faster:

In [1]: lst = range(1, 10001)



In [2]: n = 5



In [3]: %%timeit

    ...: for group in zip(*[iter(lst)] * n):

    ...:     group

    ...:

236 µs ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)



In [4]: %%timeit

    ...: for i in range(0, len(lst), n):

    ...:     lst[i:i+n]

    ...:

1.32 ms ± 184 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

edited Jan 22 at 5:48

answered Jan 22 at 5:39

Tomothy32

7,8461728

A bit more idiomatic than the other answers:

n = 5

lst = [1,2,3,4,5,6,7,8,9,10]



for group in zip(*[iter(lst)] * n):

    print(*group)

1 2 3 4 5

6 7 8 9 10

For larger lists, it's also much faster:

In [1]: lst = range(1, 10001)



In [2]: n = 5



In [3]: %%timeit

    ...: for group in zip(*[iter(lst)] * n):

    ...:     group

    ...:

236 µs ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)



In [4]: %%timeit

    ...: for i in range(0, len(lst), n):

    ...:     lst[i:i+n]

    ...:

1.32 ms ± 184 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

edited Jan 22 at 5:48

answered Jan 22 at 5:39

Tomothy32

7,8461728

edited Jan 22 at 5:48

answered Jan 22 at 5:39

Tomothy32

7,8461728

answered Jan 22 at 5:39

Tomothy32

7,8461728

answered Jan 22 at 5:39

Tomothy32

7,8461728

add a comment |

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