Question

给出向量列表：

[((0, 2.6147858445098677), (1, 1.0257184186249431)), ((0, 2.6147858445098677), (2, 0.34113605903013322)), ((0, 2.6147858445098677), (3, 0.074196986667672063)), ((1, 1.0257184186249431), (2, 0.34113605903013322)), ((1, 1.0257184186249431), (3, 0.074196986667672063)), ((2, 0.34113605903013322), (3, 0.074196986667672063))]

如何仅提取每个向量中的第一个条目？

[(0, 1), (0,2), (0, 3), (1, 2), (1,3), (2, 3)]

Answer 1

可以只使用列表推导。每个项目都是一个包含两个元组的元组，因此我们可以像这样解析每个内部元组的第一个项目：

x = [((0, 2.6147858445098677), (1, 1.0257184186249431)), ((0, 2.6147858445098677), (2, 0.34113605903013322)), ((0, 2.6147858445098677), (3, 0.074196986667672063)), ((1, 1.0257184186249431), (2, 0.34113605903013322)), ((1, 1.0257184186249431), (3, 0.074196986667672063)), ((2, 0.34113605903013322), (3, 0.074196986667672063))]
result = [(item[0][0], item[1][0]) for item in x]
print(result)

输出：

[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

根据需要。这是非常c的样式，如果需要，您可以使用pythonic并使用元组拆包：

result = [(first[0], second[0]) for  first, second in x]

这更易于理解/阅读。

Answer 2

或map：

>>> l=[((0, 2.6147858445098677), (1, 1.0257184186249431)), ((0, 2.6147858445098677), (2, 0.34113605903013322)), ((0, 2.6147858445098677), (3, 0.074196986667672063)), ((1, 1.0257184186249431), (2, 0.34113605903013322)), ((1, 1.0257184186249431), (3, 0.074196986667672063)), ((2, 0.34113605903013322), (3, 0.074196986667672063))]
>>> list(map(lambda x: (x[0][0],x[1][0]),l))
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
>>>

Answer 3

lista =[
    ((0, 2.6147858445098677), (1, 1.0257184186249431)), 
    ((0, 2.6147858445098677), (2, 0.34113605903013322)),
    ((0, 2.6147858445098677), (3, 0.074196986667672063)), 
    ((1, 1.0257184186249431), (2, 0.34113605903013322)), 
    ((1, 1.0257184186249431), (3, 0.074196986667672063)), 
    ((2, 0.34113605903013322), (3, 0.074196986667672063)) ]

new_lista = []

for i in lista:
    new_lista.append((i[0][0], i[1][0]))

print(new_lista)

或者：

[new_lista.append((i[0][0], i[1][0])) for i in lista]

(xenial)vash@localhost:~/python/stack_overflow$ python3.7 vector.py
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

Answer 4

出于完整性考虑，您还可以使用operator.itemgetter，map和functools.partial的组合（给定列表v）：

from operator import itemgetter
from functools import partial
list(map(tuple, map(partial(map, itemgetter(0)), v)))

或者在Python 2.7或更早的版本中可以更加简洁：

map(partial(map, itemgetter(0)), v)

从向量列表中，每个向量仅提取一列

4 个答案: