如何串联从python permutations（）返回的数字？

Question

I have the following code that generates all 2-digit permutations in the range 0-9:

p = permutations(range(10), 2)

Which produces a result like this:

(0, 1); (0, 2); (0, 3); (0, 4); (0, 5); (0, 6); (0, 7); (0, 8); (0, 9); (1, 0); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6); (1, 7); (1, 8); (1, 9); (2, 0); (2, 1); (2, 3); (2, 4); (2, 5); (2, 6); (2, 7); (2, 8); (2, 9); (3, 0); (3, 1); (3, 2); (3, 4); (3, 5); (3, 6); (3, 7); (3, 8); (3, 9); (4, 0); (4, 1); (4, 2); (4, 3); (4, 5); (4, 6); (4, 7); (4, 8); (4, 9); (5, 0); (5, 1); (5, 2); (5, 3); (5, 4); (5, 6); (5, 7); (5, 8); (5, 9); (6, 0); (6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 7); (6, 8); (6, 9); (7, 0); (7, 1); (7, 2); (7, 3); (7, 4); (7, 5); (7, 6); (7, 8); (7, 9); (8, 0); (8, 1); (8, 2); (8, 3); (8, 4); (8, 5); (8, 6); (8, 7); (8, 9); (9, 0); (9, 1); (9, 2); (9, 3); (9, 4); (9, 5); (9, 6); (9, 7); (9, 8)

What should I do to get an output like [01,02,03....98] and then I can get certain element by calling p[0]?

Answer 1

Just aggregate each result:

p = permutations(range(10), 2)
result = [str(x[0]) + str(x[1]) for x in p]

Answer 2

A (probably not most efficient) one-liner:

["".join([str(x) for x in elem]) for elem in p]

Or:

[("%d"*len(p[0]))%(elem) for elem in p]

Answer 3

In [13]: import itertools as IT
In [17]: p = IT.permutations(range(10), 2)

In [18]: list(IT.starmap('{}{}'.format, p))
Out[18]: 
['01',
 '02',
 '03',
 '04',
...

Answer 4

Use a list comprehension:

p = permutations(range(10), 2)
result = [ "%d%d" % (x[0], x[1]) for x in p ]

Answer 5

As John Clements and I pointed out in comments, its possible to make the strings or numbers you want without doing a permutation.

Two-permutations of digits 0 to 9 simply represent integers from 1 to 98 that don't have any repeated digits. Conveniently, all 2-digit numbers with repeated digits are multiples of 11, so you can easily make a list comprehension or generator expression that skips over the ones you don't want:

nums = [i for i in range(1, 99) if i % 11]

If you want your result to be a list of two-character strings, you use the format function with a format string that says to to pad one-digit numbers with a zero:

strings = [format(i, '02') for i in range(1, 99) if i % 11]

Unfortunately, it's a bit harder to extend this technique to longer numbers, since just skipping multiples of 11 won't cut it any more. What the best approach would be will probably depend on what you're actually doing with the values you get.

Answer 6

This works:

>>> from itertools import permutations
>>> ["%i%i" % (a,b) for a,b in permutations(range(10), 2)]
['01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '12', '13', '14', '15',
'16', '17', '18', '19', '20', '21', '23', '24', '25', '26', '27', '28', '29', '30', 
'31', '32', '34', '35', '36', '37', '38', '39', '40', '41', '42', '43', '45', '46', 
'47', '48', '49', '50', '51', '52', '53', '54', '56', '57', '58', '59', '60', '61', 
'62', '63', '64', '65', '67', '68', '69', '70', '71', '72', '73', '74', '75', '76', 
'78', '79', '80', '81', '82', '83', '84', '85', '86', '87', '89', '90', '91', '92', 
'93', '94', '95', '96', '97', '98']
>>>

Or, if you want them as integers, you can do this:

>>> from itertools import permutations
>>> [int("%i%i" % (a, b)) for a,b in permutations(range(10), 2)]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 
26, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 
49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 
72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94, 
95, 96, 97, 98]
>>>

Note however that Python does not allow you to have an integer that starts with 0.