匿名用戶

1級

2016-03-18 回答

this performs the same function as itertools.permutations

for i in range(len(l)):

s=l[:i]+l[i+1:]

These two lines can iterate all cases of permutations(order not considered) of the list's sublist of (len(list)-1)

So, like you have [1,2,3]

By this, you can iterate through:

[2,3], [1,3], [1,2]

Then you recurse it till the list is of length 1.

p = perm(s)

so, for [2,3], you get [2] and [3]

for [1,3], you get [1] and [3]

for [1,2], you get [1, 2]

Then these codes considers the order.

for x in p:

r.append(l[i:i+1]+x)

After this two processing, you can get [2,3] and [3,2] from [2,3]

[2,3] + [1] = [2,3,1]

[3,2] + [1] = [3,2,1]

So you get all cases when 2,3 is the first two of the permutations.

Well, hope this helps you.

Also, this is a equivalent function of permutations :

This returns an iterable, and I believe it is more efficient.

def permutations(iterable, r=None):

# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC

# permutations(range(3)) --> 012 021 102 120 201 210

pool = tuple(iterable)

n = len(pool)

r = n if r is None else r

if r > n:

return

indices = range(n)

cycles = range(n, n-r, -1)

yield tuple(pool[i] for i in indices[:r])

while n:

for i in reversed(range(r)):

cycles[i] -= 1

if cycles[i] == 0:

indices[i:] = indices[i+1:] + indices[i:i+1]

cycles[i] = n - i

else:

j = cycles[i]

indices[i], indices[-j] = indices[-j], indices[i]

yield tuple(pool[i] for i in indices[:r])

break

else:

return

追問：

Then these codes considers the order.

order 是怎么產(chǎn)生的,麻煩再回答一下啦

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