考虑
a = [1,2,3,4]
i = 0
j = 1
for i in range(len(a)):
for j in range(len(a)):
d = (a[i]-a[j])
j = j + 1
print i, j, d
i = i + 1
输出
0 1 0
0 2 -1
0 3 -2
0 4 -3
1 1 1
1 2 0
1 3 -1
1 4 -2
2 1 2
2 2 1
2 3 0
2 4 -1
3 1 3
3 2 2
3 3 1
3 4 0
我正在尝试遍历我的数组,这样我才能得到d为非零的数字,并且我不会重复相同的i和j(例如:如果i = 0,j = 1或i = 1,j = 0)。它就像做一个组合问题,我正在寻找我的数组中的对数和它的d。
答案 0 :(得分:1)
只需使用permutations
中的itertools
:
import itertools
a = [1,2,3,4]
for permutation in itertools.permutation(a, 2):
print permutation
输出
(1, 2)
(1, 3)
(1, 4)
(2, 1)
(2, 3)
...
...
如果您还想要距离
a = [1,2,3,4]
for permutation in itertools.permutation(a, 2):
print permutation, permutation[1] - permutation[0]
(1, 2) 1
(1, 3) 2
(1, 4) 3
(2, 1) -1
答案 1 :(得分:1)
我正在尝试遍历我的数组,以便我只能获得数字 d
非零
除非这是家庭作业,否则我建议您使用itertools.combinations反向排序列表或itertools.permutations来解决您的问题
>>> list((a,b) for a,b in itertools.permutations(a, 2) if a > b)
[(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)]
>>> list(itertools.combinations(sorted(a, reverse = True), 2))
[(4, 3), (4, 2), (4, 1), (3, 2), (3, 1), (2, 1)]
答案 2 :(得分:0)
试试这个:
a = [1,2,3,4]
i = 0
j = 1
for i in range(len(a)):
for j in range(len(a)):
d = (a[i]-a[j])
j = j + 1
if i != j and d != 0:
print i, j, d
i = i + 1
输出:
>>>
0 2 -1
0 3 -2
0 4 -3
1 3 -1
1 4 -2
2 1 2
2 4 -1
3 1 3
3 2 2