使用Python中的嵌套循环计算所有唯一排列

Question

What would be the python equivalent implementation for this C++ code:

char x[10];
for (int i=0; i < 10; i++) {
    for (int j=i; j < 10; j++) {
        calc_something(x[i], x[j])
    }
}

Thank you

Answer 1

This is done simply with itertools.combinations() :

import itertools

...

for i, j in itertools.combinations(x, 2):
    calc_something(i, j)

This gives what you want. Specifically, it will return the elements in this order:

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

Answer 2

Here are some solutions that don't use imports, and assuming that x is already declared as a list containing 10 elements:

for i in range(10):  # xrange in Python 2
    for j in range(i, 10):
        calc_something(x[i], x[j])

Or using the enumerate function:

for i, el in enumerate(x):
    for j in x[i:]:
        calc_something(el, j)

Answer 3

simplest would be :

    for i in range(1,10):
        for j in range(1,10):
           calc_something( list[i],list[j])

instead of hard coding (1,10) you can say
for i in list: for j in list:

Answer 4

x=[]

for i in range(1,10):
    for j in range(1,10):
        calc_something(x[i],x[j])