Question

给出了几个清单：

a = ["a1", "a2", "a3"]
b = ["b1", "b2", "b3"]
...
n = ["n1", "n2", "n3"]

以及新值列表：

new_vals = ["a4", "b4", "n4"]

我想得到：

["a1", "a2", "a3", "a4"]
["b1", "b2", "b3", "b4"]
...
["n1", "n2", "n3", "n4"]

当然，我可以使用循环和临时变量来完成此操作。似乎zip，map和list.extend的组合应该更优雅地完成，但是我不知道。

Answer 1

这样的事情：

a = ["a1", "a2", "a3"]
b = ["b1", "b2", "b3"]

# Put the list a, b ... in a big_list. 
big_list = [a, b]

new_vals = ["a4", "b4", "n4"]

for i, new_val in enumerate(new_vals):
    big_list[i].append(new_val)

Answer 2

假设您有一个列表列表：

lsts = [ ['a1','a2','a3'],
         ['b1','b2','b3'],
         ['c1','c2','c3'] ]

以及一个列表，其中包含要追加到lsts中每个列表末尾的新值：

lst = [ 'a4', 'b4', 'c4' ]

然后你可以使用列表理解：

new_lsts = [l + [x] for l, x in zip(lsts, lst)]

Answer 3

面向map（）的解决方案：

a = ["a1", "a2", "a3"]
b = ["b1", "b2", "b3"]
n = ["n1", "n2", "n3"]
new_vals = ["a4", "b4", "n4"]

map(lambda lst, x: lst.append(x), (a, b, n), new_vals)

并合并上面的列表建议列表：

lsts = [['a1','a2','a3'],
        ['b1','b2','b3'],
        ['c1','c2','c3']]
new_vals = ["a4", "b4", "c4"]
map(lambda lst, x: lst.append(x), lsts, new_vals)

这可能更好，因为它可以就地修改lsts而不是创建新的列表列表。

优雅地扩展几个列表

3 个答案: