Answers:
可以使用列表推导来完成。这些基本上采取的形式[function-of-item for item in some-list]
。例如,要创建一个新列表,其中所有项目均为小写(或在第二个片段中为大写),则可以使用:
>>> [x.lower() for x in ["A","B","C"]]
['a', 'b', 'c']
>>> [x.upper() for x in ["a","b","c"]]
['A', 'B', 'C']
您还可以使用以下map
功能:
>>> map(lambda x:x.lower(),["A","B","C"])
['a', 'b', 'c']
>>> map(lambda x:x.upper(),["a","b","c"])
['A', 'B', 'C']
list(map(lambda x:x.upper(),["a","b","c"]))
map(str.lower, ["A","B","C"])
除了易于阅读(对于许多人而言)之外,列表理解还赢得了速度竞赛:
$ python2.6 -m timeit '[x.lower() for x in ["A","B","C"]]'
1000000 loops, best of 3: 1.03 usec per loop
$ python2.6 -m timeit '[x.upper() for x in ["a","b","c"]]'
1000000 loops, best of 3: 1.04 usec per loop
$ python2.6 -m timeit 'map(str.lower,["A","B","C"])'
1000000 loops, best of 3: 1.44 usec per loop
$ python2.6 -m timeit 'map(str.upper,["a","b","c"])'
1000000 loops, best of 3: 1.44 usec per loop
$ python2.6 -m timeit 'map(lambda x:x.lower(),["A","B","C"])'
1000000 loops, best of 3: 1.87 usec per loop
$ python2.6 -m timeit 'map(lambda x:x.upper(),["a","b","c"])'
1000000 loops, best of 3: 1.87 usec per loop
map
赢得比赛,但无能为力:)
map(str.lower,["A","B","C"])
最快的是python3.7.5
列表理解是我要做的,这是“ Pythonic”的方式。以下记录显示了如何将列表转换为全部大写然后转换回小写:
pax@paxbox7:~$ python3
Python 3.5.2 (default, Nov 17 2016, 17:05:23)
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> x = ["one", "two", "three"] ; x
['one', 'two', 'three']
>>> x = [element.upper() for element in x] ; x
['ONE', 'TWO', 'THREE']
>>> x = [element.lower() for element in x] ; x
['one', 'two', 'three']
list
作为变量名是不是最好的选择:)
list
东西:))。您认为您最近得到的紫外线来自哪里?:)
对于此样本,理解最快
$ python -m timeit -s's = [“一个”,“两个”,“三个”] * 1000''[x.s中x的x.upper] 1000次循环,最好是3次:每个循环809微秒 $ python -m timeit -s's = [“一个”,“两个”,“三个”] * 1000''map(str.upper,s)' 1000次循环,每循环3:1.12毫秒的最佳时间 $ python -m timeit -s's = [“一个”,“两个”,“三个”] * 1000''map(lambda x:x.upper(),s)' 1000个循环,每个循环最好3:1.77毫秒
解:
>>> s = []
>>> p = ['This', 'That', 'There', 'is', 'apple']
>>> [s.append(i.lower()) if not i.islower() else s.append(i) for i in p]
>>> s
>>> ['this', 'that', 'there', 'is','apple']
此解决方案将创建一个单独的列表,其中包含小写字母的项目,无论它们的原始大小写如何。如果原始大小写为大写,则中的list s
会包含相应项目的小写list p
。如果列表项的原始大小写已经为小写,list p
则list s
它将保留该项目的大小写并将其保持为小写。现在,您可以使用list s
代替list p
。
如果您的目的是通过一次转换来与另一个字符串匹配,则也可以使用str.casefold()
。
(:集体VS集体EG).Though当你有非ASCII字符和匹配使用ASCII版本,这是有用str.lower
还是str.upper
在这种情况下出现故障时,str.casefold()
会通过。这在Python 3中可用,并且通过回答https://stackoverflow.com/a/31599276/4848659详细讨论了这个想法。
>>>str="Hello World";
>>>print(str.lower());
hello world
>>>print(str.upper());
HELLO WOLRD
>>>print(str.casefold());
hello world
的Python3.6.8
In [1]: a = 'which option is the fastest'
In [2]: %%timeit
...: ''.join(a).upper()
762 ns ± 11.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [3]: %%timeit
...: map(lambda x:x.upper(), a)
209 ns ± 5.73 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [4]: %%timeit
...: map(str.upper, [i for i in a])
1.18 µs ± 11.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [5]: %%timeit
...: [i.upper() for i in a]
3.2 µs ± 64.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
如果您需要一个字符串或列表作为输出而不是一个迭代器(这是针对Python3的),则可以使用compare ''.join(string).upper()
选项:
In [10]: %%timeit
...: [i for i in map(lambda x:x.upper(), a)]
4.32 µs ± 112 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
如果您尝试将列表中的所有字符串都转换为小写,则可以使用pandas:
import pandas as pd
data = ['Study', 'Insights']
pd_d = list(pd.Series(data).str.lower())
输出:
['study', 'insights']