在 python 中打印时，在特定数量的输出后创建新行

Question

Suppose there is a list; a = [1, 2, 3, 4, 5, 6, 7, 8, 9]

Now, if I use print(*a), I'll get

1 2 3 4 5 6 7 8 9

But, I want to print it as;

1 2 3
4 5 6
7 8 9

Here, the key point is making a new line after specific numbers of outputs , not after specific outputs .

Any built-in functions to achieve this? If not, I'm open to hearing alternative ways.

Answer 1

You can use modulo operator with integer 3

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

for i in a:
    if i%3:
        print(i, end =" ")
    else:
        print(i)

if those integer members are quoted such as a=['1','2',"3",...] , then use casting to int as by replacing i%3 with int(i)%3

or the members are completely arbitrary string values such as a = ['1a', '4', "2p", ...] , then use and index( j ) for iterayion such as

j=0
for i in a:
    j+=1 
    if j%3:
        print(i, end =" ")
    else:
        print(i)

Answer 2

Your problem is a bit too special for there to be a general solution in the print function. There is pprint for pretty-printing data structures, but as far as I can see even that does not have this option.

So this might be the shortest way. The function takes n elements from the array and prints them each.

from math import ceil


def printN(n, array):
    for i in range(ceil(len(array) / n)):
        print(*array[i*n:(i+1)*n])

array = [1, 2, 3, 4, 5, 6, 7, 8, 9]
printN(3, array)