使用場景
列舉組合:
問題是這樣的.
有 n 個列表,分別從每個列表中取出乙個元素,一共有多少種組合?
例如:
a = ['a1','a2']
b = ['b1','b2','b3']
[
('a1','b1'),
('a1','b2'),
('a1','b3'),
('a2','b1'),
('a2','b2'),
('a2','b3')
]
'''
'''a = ['a1','a2']
b = ['b1','b2','b3']
res =
for i in a:
for j in b:
print(res)
a = ['a1','a2']
b = ['b1','b2','b3']
res =
for i in a:
for j in b:
for k in c:
......
我們可以先將第乙個和第二個組合,再拿組合出的結果和第三個組合,依次類推...
如下如所示:
用**表示如下:
迭代
def merge(i,j):
"""i = "a"
j = ("b","c")
return: ("a","b","c")
"""res =
for p in (i,j):
if isinstance(p,tuple):
res.extend(p)
else:
return tuple(res)
def combinen(*args):
target = args[0]
for li in args[1:]:
tmp =
for i in target:
for j in li:
target = tmp
return target
def merge(i,j):
"""i = "a"
j = ("b","c")
return: ("a","b","c")
"""res =
for p in (i,j):
if isinstance(p,tuple):
res.extend(p)
else:
return tuple(res)
def combine2(a, b):
res =
for i in a:
for j in b:
return res
def combinenrecursion(*args):
if len(args) == 2:
return combine2(*args)
return combine2(args[0],combinenrecursion(*args[1:]))
def combinen(*li):
res =
# 相當於最內層迴圈執行的次數.
total_times = reduce(lambda x, y: x*y, [len(item) for item in li])
n = 0
while n < total_times:
tmp = n
tem_res =
for i in range(len(li)):
# 餘數就是參與計算的元素的下標,商用於計算下乙個列表參與元素的下標.
tmp, cur = divmod(tmp, len(li[i]))
n += 1
return res
res = combinen(["a1","a2"], ["b1", "b2"], ["c1", "c2"])
for i in res:
print(i)
['a1', 'b1', 'c1']
['a2', 'b1', 'c1']
['a1', 'b2', 'c1']
['a2', 'b2', 'c1']
['a1', 'b1', 'c2']
['a2', 'b1', 'c2']
['a1', 'b2', 'c2']
['a2', 'b2', 'c2']
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