[英]Get a list with tuple unpacking in a function
我有一個給定的 function 接受不同的輸入(示例):
def myfunction(x, y, z):
a = x,y,z
return a
然后,這個for循環:
tripples = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'm')]
for tripple in tripples:
lst.append(myfunction(*tripple))
lst
像這樣工作:
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'm')]
我想i in range(n)
運行它並獲取列表列表作為 output,
for i in range(3):
for tripple in tripples:
lst_lst.append(myfunction(*tripple))
lst_lst
Output:
[('a', 'b', 'c'),
('d', 'e', 'f'),
('g', 'h', 'i'),
('j', 'k', 'm'),
('a', 'b', 'c'),
('d', 'e', 'f'),
('g', 'h', 'i'),
('j', 'k', 'm'),
('a', 'b', 'c'),
('d', 'e', 'f'),
('g', 'h', 'i'),
('j', 'k', 'm')]
所需的 output:
[[('a', 'b', 'c'),
('d', 'e', 'f'),
('g', 'h', 'i'),
('j', 'k', 'm')],
[('a', 'b', 'c'),
('d', 'e', 'f'),
('g', 'h', 'i'),
('j', 'k', 'm')],
[('a', 'b', 'c'),
('d', 'e', 'f'),
('g', 'h', 'i'),
('j', 'k', 'm')]]
如果它有幫助,完整的代碼:
def myfunction(x, y, z):
a = x,y,z
return a
lst = []
lst_lst = []
tripples = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'm')]
for tripple in tripples:
lst.append(myfunction(*tripple))
for i in range(3):
for tripple in tripples:
lst_lst.append(myfunction(*tripple))
lst_lst
def myfunction(x, y, z):
a = x,y,z
return a
lst = []
tripples = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'm')]
for i in range(3):
lst_lst = []
for tripple in tripples:
lst_lst.append(myfunction(*tripple))
lst.append(lst_lst)
print(lst)
您需要使用一個臨時列表,它保存一個循環的結果,然后將這些結果添加到最終列表中,並在下一個循環中初始化自身,然后再次保存下一個三元組的結果
def myfunction(x, y, z):
a = x,y,z
return a
lst = []
lst_lst = []
tripples = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'm')]
for tripple in tripples:
lst.append(myfunction(*tripple))
for i in range(3):
tmp =[]
for tripple in tripples:
tmp.append(myfunction(*tripple))
lst_lst.append(tmp)
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