在Python中打印不带方括号的元组列表

Question

I am attempting to print out a list of tuples without square brackets whilst maintaining parentheses around the tuple.

For example, instead of the output being (current output):

[('1', '3'), ('1', '4'), ('2', '3'), ('2', '4')]

The output should be:

(1, 3) (1, 4) (2, 3) (2, 4)

Current source code below.

from itertools import product

if __name__ == '__main__':
    input_lists = []
    num_loops = int(input())
    i = 0

    while i < num_loops:
        add_to_list = input()
        input_lists.append(add_to_list.split(" "))
        i += 1

    result = ([i for i in product(*input_lists)])
    print(''.join(str(result)))

Answer 1

List comprehension, str.format and str.join :

In [1045]: lst = [('1', '3'), ('1', '4'), ('2', '3'), ('2', '4')]

In [1046]: ' '.join('({}, {})'.format(i, j) for i, j in lst)
Out[1046]: '(1, 3) (1, 4) (2, 3) (2, 4)'

Answer 2

I suggest the int conversion and then unpack:

>>> from __future__ import print_function  # for Python 2
>>> lst = [('1', '3'), ('1', '4'), ('2', '3'), ('2', '4')]
>>> print(*[tuple(map(int,t)) for t in lst])
(1, 3) (1, 4) (2, 3) (2, 4)

Answer 3

print(' '.join([str(int(i for i in tup)
                for tup in list_of_tuples]))

Calling str() on a tuple produces the tuple itself, really, so I just did this for each item in the list. I also needed to make each item in the tuple an int so I also performed int() on each item in each tuple. The ' '.join() method will separate all of the items in the iterable passed in by a single whitespace. So... I passed in a list comprehension which performs str() on each item in the list.

Answer 4

With an array like

 x = [('1', '3'), ('1', '4'), ('2', '3'), ('2', '4')]

you can do simply:

l = [(int(x),int(y)) for x,y in l]
print(*l)

Its output is similar to

output//(1, 3) (1, 4) (2, 3) (2, 4)

Answer 5

Here's my take:

print(' '.join([str(tuple(map(int, i))) for i in L]))

A couple of points:

• We use tuple with map to convert values from str to int .
• str.join is one of the few instances where passing a list is more efficient than generator comprehension.

Answer 6

You can unpack the list.

lst = [(1, 2), (3, 4)]
print(*lst)

Output:

(1, 2) (3, 4)