将元组列表划分为多个列表的更简洁方法

时间:2021-07-15 08:30:59

标签: python numpy

标题是不言自明的,是否有单行表达?

可重现的例子:

Data = [(21,1,2000), (30,0,1991), (27,0,1994)]

age = [el[0] for el in Data]
sex = [el[1] for el in Data]
birth = [el[2] for el in Data]

print(age, sex, birth)

3 个答案:

答案 0 :(得分:7)

使用zip

age, sex, birth = zip(*Data)  # tuple

# OR

age, sex, birth = map(list, zip(*Data))  # list
>>> age
[21, 30, 27]

>>> sex
[1, 0, 0]

>> birth
[2000, 1991, 1994]

答案 1 :(得分:3)

对@Corralien 解决方案稍作修改,要在list 中而不是在tuple 中获得结果,将mapzip 一起使用以获得结果

Data = [(21,1,2000), (30,0,1991), (27,0,1994)]
age, sex, birth = map(list, zip(*Data))
print(age, sex, birth, sep ='\n')

输出

[21, 30, 27] # age
[1, 0, 0] # sex
[2000, 1991, 1994] # birth

答案 2 :(得分:0)

# You can do this by combining zip() with map() :

Data = [(21,1,2000), (30,0,1991), (27,0,1994)]
result = list(map(list, zip(*Data)))
print(result) 
# [[21, 30, 27], [1, 0, 0], [2000, 1991, 1994]]
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