如何生成列表的所有排列?例如:
permutations([])
[]
permutations([1])
[1]
permutations([1, 2])
[1, 2]
[2, 1]
permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
如何生成列表的所有排列?例如:
permutations([])
[]
permutations([1])
[1]
permutations([1, 2])
[1, 2]
[2, 1]
permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
当前回答
#!/usr/bin/env python
def perm(a, k=0):
if k == len(a):
print a
else:
for i in xrange(k, len(a)):
a[k], a[i] = a[i] ,a[k]
perm(a, k+1)
a[k], a[i] = a[i], a[k]
perm([1,2,3])
输出:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 2, 1]
[3, 1, 2]
当我交换列表的内容时,需要一个可变的序列类型作为输入。例如,烫发(list(“ball”)会起作用,而烫发(“ball”)不会起作用,因为你不能更改字符串。
这种Python实现的灵感来自Horowitz、Sahni和Rajasekeran在《计算机算法》一书中提出的算法。
其他回答
生成所有可能的排列
我正在使用python3.4:
def calcperm(arr, size):
result = set([()])
for dummy_idx in range(size):
temp = set()
for dummy_lst in result:
for dummy_outcome in arr:
if dummy_outcome not in dummy_lst:
new_seq = list(dummy_lst)
new_seq.append(dummy_outcome)
temp.add(tuple(new_seq))
result = temp
return result
测试用例:
lst = [1, 2, 3, 4]
#lst = ["yellow", "magenta", "white", "blue"]
seq = 2
final = calcperm(lst, seq)
print(len(final))
print(final)
这是受Haskell实现使用列表理解的启发:
def permutation(list):
if len(list) == 0:
return [[]]
else:
return [[x] + ys for x in list for ys in permutation(delete(list, x))]
def delete(list, item):
lc = list[:]
lc.remove(item)
return lc
免责声明:无耻的插件由包作者。:)
trotter包与大多数实现的不同之处在于,它生成的伪列表实际上不包含排列,而是描述排列与排序中各个位置之间的映射,从而可以处理非常大的排列“列表”,如本演示所示,它在一个包含字母表中所有字母排列的伪列表中执行相当即时的操作和查找,而不使用比典型网页更多的内存或处理。
在任何情况下,要生成排列列表,我们可以执行以下操作。
import trotter
my_permutations = trotter.Permutations(3, [1, 2, 3])
print(my_permutations)
for p in my_permutations:
print(p)
输出:
A pseudo-list containing 6 3-permutations of [1, 2, 3]. [1, 2, 3] [1, 3, 2] [3, 1, 2] [3, 2, 1] [2, 3, 1] [2, 1, 3]
我的Python解决方案:
def permutes(input,offset):
if( len(input) == offset ):
return [''.join(input)]
result=[]
for i in range( offset, len(input) ):
input[offset], input[i] = input[i], input[offset]
result = result + permutes(input,offset+1)
input[offset], input[i] = input[i], input[offset]
return result
# input is a "string"
# return value is a list of strings
def permutations(input):
return permutes( list(input), 0 )
# Main Program
print( permutations("wxyz") )
def permuteArray (arr):
arraySize = len(arr)
permutedList = []
if arraySize == 1:
return [arr]
i = 0
for item in arr:
for elem in permuteArray(arr[:i] + arr[i + 1:]):
permutedList.append([item] + elem)
i = i + 1
return permutedList
我不打算在一个新的行中穷尽所有的可能性,以使它有点独特。