如何在新行上打印python列表的第n个索引？

Question

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!

Answer 1

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

Answer 2

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.

Answer 3

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)

Answer 4

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.