通过将字符串与匹配的索引值连接来组合列表

Question

我有两个要合并的列表，但实际上我希望加入具有匹配索引的项目，而不是增加列表中的项目数。例如：

List1 = ['A', 'B', 'C']
List2 = ['1', '2', '3']
List3 = ['A1', 'B2', 'C3']

我已经看到了很多关于简单地组合两个列表的问题，但是恐怕我还没有找到任何可以实现的东西。

任何帮助都将不胜感激。干杯。

Answer 1

map(''.join,zip(List1,List2,List3))
>>> 
['A1A1', 'B2B2', 'C3C3']

---

说明：

zip(List1,List2,List3)

回报

[('A', '1', 'A1'), ('B', '2', 'B2'), ('C', '3', 'C3')]

每个元组都会重复压缩列表中与索引N关联的元素。我们想把每个元组中的元素组合成一个字符串。对于一个元组，我们可以：

>>> ''.join(('A', '1', 'A1'))
'A1A1'

为了产生所需的结果，从而获得所有所需字符串的列表，我们将这个join函数映射到所有元组，如下所示：

map(''.join,zip(List1,List2,List3))

导致

['A1A1', 'B2B2', 'C3C3']

---

所以如果你只想添加List1和List2

map(''.join,zip(List1,List2))
>>> 
['A1', 'B2', 'C3']

一些时间：

Microsoft Windows [Version 6.2.9200]
(c) 2012 Microsoft Corporation. All rights reserved.

C:\Users\Henry>python -m timeit -s "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2','3']*10**5" "map(lambda x, y: x + y, List1, List2)"
10 loops, best of 3: 44 msec per loop

C:\Users\Henry>python -m timeit -s "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2', '3']*10**5" "[x + y for x, y in zip(List1, List2)]"
10 loops, best of 3: 44 msec per loop

C:\Users\Henry>python -m timeit -s "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2', '3']*10**5" "map(''.join,zip(List1,List2))"
10 loops, best of 3: 42.6 msec per loop

C:\Users\Henry>python -m timeit -s "from operator import add" "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2', '3']*10**5" "map(add, List1, List2)"
10 loops, best of 3: 28.6 msec per loop

使用izip而不是zip

Microsoft Windows [Version 6.2.9200]
(c) 2012 Microsoft Corporation. All rights reserved.

C:\Users\Henry>python -m timeit -s "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2','3']*10**5" "map(lambda x, y: x + y, List1, List2)"
10 loops, best of 3: 44.1 msec per loop

C:\Users\Henry>python -m timeit -s "from itertools import izip" "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2', '3']*10**5" "[x + y for x, y in izip(List1, List2)]"
10 loops, best of 3: 31.3 msec per loop

C:\Users\Henry>python -m timeit -s "from itertools import izip" "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2', '3']*10**5" "map(''.join,izip(List1,List2))"
10 loops, best of 3: 36.2 msec per loop

C:\Users\Henry>python -m timeit -s "from operator import add" "List1 = ['A', 'B', 'C']*10**5; List2 = ['1', '2', '3']*10**5" "map(add, List1, List2)"
10 loops, best of 3: 28.6 msec per loop

Answer 2

>>> List1 = ['A', 'B', 'C']
>>> List2 = ['1', '2', '3']
>>> [x + y for x, y in zip(List1, List2)]
['A1', 'B2', 'C3']

Answer 3

>>> List1 = ['A', 'B', 'C']
>>> List2 = ['1', '2', '3']
>>> map(lambda a, b: a + b, List1, List2)
['A1', 'B2', 'C3']