CODEWITHSUNDEEP
python
Smple_V
Smple_V

Reputation: 868

how to print 3x3 array in python?

I need to print a 3 x 3 array for a game called TicTackToe.py. I know we can print stuff from a list in a horizontal or vertical way by using 

listA=['a','b','c','d','e','f','g','h','i','j']
# VERTICAL PRINTING 
for item in listA:
        print item

Output:

a
b
c

or

# HORIZONTAL  PRINTING
for item in listA:
        print item,

Output: 

a b c d e f g h i j

How can I print a mix of both, e.g. printing a 3x3 box like 

a b c
d e f
g h i

Upvotes: 16

Views: 13552

Answers (8)

Gopal Sankar
Gopal Sankar

Reputation: 11

An easy solution #The end="" will print the values in the same line and the empty print() will make the control move to the immediately next line. #Don't use "\n", for it will create an empty line between the rows!

s = []
for i in range(3):
    s.append(list(map(int, input().split())))
for i in range(3):
    for j in range(3):
        print(s[i][j],"",end="")
    print()

Upvotes: 0

Miles Gillham
Miles Gillham

Reputation: 294

print ''.join(' '*(n%3!=0)+l+'\n'*(n%3==2) for n,l in enumerate(listA))

a b c
d e f
g h i
j

Python is awesome.

Upvotes: 0

Open AI - Opting Out
Open AI - Opting Out

Reputation: 24133

You can enumerate the items, and print a newline only every third item:

for index, item in enumerate('abcdefghij', start=1):
    print item,
    if not index % 3:
        print

Output:

a b c
d e f
g h i
j

enumerate starts counting from zero by default, so I set start=1.

As @arekolek comments, if you're using Python 3, or have imported the print function from the future for Python 2, you can specify the line ending all in one go, instead of the two steps above:

for index, item in enumerate('abcdefghij', start=1):
    print(item, end=' ' if index % 3 else '\n')

Upvotes: 23

Ian
Ian

Reputation: 30813

One method which have not been mentioned: Get the slice of the list mathematically and print by the slice of the list.

In Python 2.x:

listA=['a','b','c','d','e','f','g','h','i','j']
val = (len(listA) + 2) // 3
for i in range(val):
    print(' '.join(listA[i*3:(i+1)*3]))

In Python 3.x:

listA=['a','b','c','d','e','f','g','h','i','j']
val = (len(listA) + 2) // 3
for i in range(val):
    print(*listA[i*3:(i+1)*3], sep=" ")

Upvotes: 1

Mad Physicist
Mad Physicist

Reputation: 114320

You could use the format method of the string object:

for i in range(3):
    print "{} {} {}".format(*listA[3*i:3*i+3])

Also, instead of multiplying the index by 3 at every iteration, you can just take steps of three elements through the list:

for i in range(0, len(listA), 3):
    print "{} {} {}".format(*listA[i:i+3])

Upvotes: 4

Padraic Cunningham
Padraic Cunningham

Reputation: 180411

You can use the logic from the grouper recipe:

listA=['a','b','c','d','e','f','g','h','i','j']

print("\n".join(map(" ".join, zip(*[iter(listA)] * 3))))
a b c
d e f
g h i

If you don't want to lose elements use izip_longest with an empty string as a fillvalue:

listA=['a','b','c','d','e','f','g','h','i','j']
from itertools import izip_longest

print("\n".join(map(" ".join, izip_longest(*[iter(listA)] * 3,fillvalue=""))))

Which differs in that it keeps the j:

a b c
d e f
g h i
j

You can put the logic in a function and call it when you want to print, passing in whatever values you want.

from itertools import izip_longest

def print_matrix(m,n, fill):
    print( "\n".join(map(" ".join, izip_longest(*[iter(m)] * n, fillvalue=fill))))

Or without itertools just chunk and join, you can also take a sep arg to use as the delimiter:

def print_matrix(m,n, sep):
    print( "\n".join(map("{}".format(sep).join, (m[i:i+n] for i in range(0, len(m), n)))))

You just need to pass the list and the size for each row:

In [13]: print_matrix(listA, 3, " ")
a b c
d e f
g h i
j

In [14]: print_matrix(listA, 3, ",")
a,b,c
d,e,f
g,h,i
j

In [15]: print_matrix(listA, 4, ",")
a,b,c,d
e,f,g,h
i,j

In [16]: print_matrix(listA, 4, ";")
a;b;c;d
e;f;g;h
i;j

Upvotes: 11

en_Knight
en_Knight

Reputation: 5381

An alternative to the two answers given here is to use something that does the formatting for you, like numpy. This is an external dependency you might not want in general, but if you're already storing things in grids/matrices, it can be a logical choice:

from numpy import array
str(array(listA).reshape((3,3)))

Put it in an array, reshape it in your favourite (compatible) shape, and make it a string. Couldn't be more intuitive!

Upvotes: 5

Maecky
Maecky

Reputation: 2038

A simple approach would be to use the modulo operator:

listA=['a','b','c','d','e','f','g','h','i','j']

count = 0

for item in listA:
    if not count % 3:
        print
    print item,
    count += 1

As pointed out by Peter Wood, you can use the enumerator operator, to avoid the count variable:

listA=['a','b','c','d','e','f','g','h','i','j']

listB = enumerate(listA)
for item in listB:
    if not item[0] % 3:
        print
    print item[1],

Upvotes: 8

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