迭代python列表（另一個列表中第二個項目的列表）

Question

我有以下形式的清單：

[['id_5', 4], ['id_6', 4], ['id_7', 4], ['id_0', 12], ['id_1', 4], ['id_2', 8], ['id_3', 8], ['id_4', 4], ['id_8', 1]]
[['id_5', 5], ['id_6', 5], ['id_7', 5], ['id_0', 15], ['id_1', 5], ['id_2', 10], ['id_3', 10], ['id_4', 5]]

我只想要每個列表的第二個元素。

所需的輸出：

[4,4,4,12,4,8,8,4,1]
[5,5,5,15,5,10,10,5]

要么：

4 4 4 12 4 8 8 4 1 
5 5 5 15 5 10 10 5 

我嘗試了這段代碼：

for i in range(0,len(vectorized)):
    for j in range (0,len(vectorized[i])):
        print(vectorized[i][j][1])

但是我的輸出是：

4
4
4
12
4
...

謝謝。

Answer 1

嘗試列表理解：

list1 = [['id_5', 4], ['id_6', 4], ['id_7', 4], ['id_0', 12], ['id_1', 4], ['id_2', 8], ['id_3', 8], ['id_4', 4], ['id_8', 1]]
list2 = [['id_5', 5], ['id_6', 5], ['id_7', 5], ['id_0', 15], ['id_1', 5], ['id_2', 10], ['id_3', 10], ['id_4', 5]]
list1_out = [i[1] for i in list1]
list2_out = [i[1] for i in list2]

輸出 ：

[4, 4, 4, 12, 4, 8, 8, 4, 1] # list1_out
[5, 5, 5, 15, 5, 10, 10, 5] # list2_out

Answer 2

有不同的方法，簡單的理解就可以做到：

[x[1] for x in l]

同樣與map和operator.itemgetter ：

from operator import itemgetter
list(map(itemgetter(1), l))

為了打印結果，您可以打開包裝進行打印：

print(*out_list)

例如：

>>> l = [['id_5', 4], ['id_6', 4], ['id_7', 4], ['id_0', 12], ['id_1', 4], ['id_2', 8], ['id_3', 8], ['id_4', 4], ['id_8
', 1]]
>>> from operator import itemgetter
>>> out_list = list(map(itemgetter(1), l))
>>> out_list
[4, 4, 4, 12, 4, 8, 8, 4, 1]
>>> print(*out_list)
4 4 4 12 4 8 8 4 1

全部作為一個班輪：

>>> print(*(x[1] for x in l))
4 4 4 12 4 8 8 4 1

Answer 3

嘗試這個：

a = [['id_5', 4], ['id_6', 4], ['id_7', 4], ['id_0', 12], ['id_1', 4], ['id_2', 8], ['id_3', 8], ['id_4', 4], ['id_8', 1]]
b = [['id_5', 5], ['id_6', 5], ['id_7', 5], ['id_0', 15], ['id_1', 5], ['id_2', 10], ['id_3', 10], ['id_4', 5]]
x = list(list(zip(*a))[1])
y = list(list(zip(*b))[1])
print(x)
print(y)

輸出：-

[4, 4, 4, 12, 4, 8, 8, 4, 1]
[5, 5, 5, 15, 5, 10, 10, 5]

Answer 4

除非我遺漏了一些東西，否則您會過於復雜。 只需遍歷主列表並打印在每個循環中獲得的列表的第二個元素，例如：

for i in big_list:
    print i[1]  

Python有一些遍歷列表的非常好的方法，因此您不需要使用range方法。