Question

我有一个字符串列表，我想知道列表中最长字符串的长度。有一个简单的方法来获得它吗？

更一般地说，我经常想知道数据框中列的最长字符串的长度。我只需要了解数据是什么样的，所以我希望有一个方便的方法，如df['column'].maxlength，而不是去for loop来获取数字。

Answer 1

这是比较：

#!/usr/bin/python

import cProfile
from timeit import Timer
from faker import Faker

def longest1(lists):
    return max(len(s) for s in lists)

def longest2(lists):
    return len(max(lists, key=len))

def longest3(lists):
    return len(sorted(lists, key=len)[-1])

s = Faker()
seq = [ s.word() for x in range(100) ]

func = [ longest1, longest2, longest3 ]

for f in func:
    t = Timer(lambda: f(seq))
    print f.__name__, cProfile.run('t.timeit(number=1000)')

结果：longest2是最快的

输出：

longest1          204011 function calls in 0.046 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.046    0.046 <string>:1(<module>)
     1000    0.000    0.000    0.045    0.000 long.py:22(<lambda>)
     1000    0.001    0.000    0.045    0.000 long.py:7(longest1)
   101000    0.025    0.000    0.031    0.000 long.py:8(<genexpr>)
        1    0.000    0.000    0.000    0.000 timeit.py:143(setup)
        1    0.000    0.000    0.046    0.046 timeit.py:178(timeit)
        1    0.000    0.000    0.046    0.046 timeit.py:96(inner)
        1    0.000    0.000    0.000    0.000 {gc.disable}
        1    0.000    0.000    0.000    0.000 {gc.enable}
        1    0.000    0.000    0.000    0.000 {gc.isenabled}
        1    0.000    0.000    0.000    0.000 {globals}
   100000    0.007    0.000    0.007    0.000 {len}
     1000    0.013    0.000    0.044    0.000 {max}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
        2    0.000    0.000    0.000    0.000 {time.time}


None
longest2          4011 function calls in 0.011 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.011    0.011 <string>:1(<module>)
     1000    0.001    0.000    0.010    0.000 long.py:10(longest2)
     1000    0.000    0.000    0.011    0.000 long.py:22(<lambda>)
        1    0.000    0.000    0.000    0.000 timeit.py:143(setup)
        1    0.000    0.000    0.011    0.011 timeit.py:178(timeit)
        1    0.000    0.000    0.011    0.011 timeit.py:96(inner)
        1    0.000    0.000    0.000    0.000 {gc.disable}
        1    0.000    0.000    0.000    0.000 {gc.enable}
        1    0.000    0.000    0.000    0.000 {gc.isenabled}
        1    0.000    0.000    0.000    0.000 {globals}
     1000    0.000    0.000    0.000    0.000 {len}
     1000    0.010    0.000    0.010    0.000 {max}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
        2    0.000    0.000    0.000    0.000 {time.time}


None
longest3          4011 function calls in 0.031 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.031    0.031 <string>:1(<module>)
     1000    0.001    0.000    0.031    0.000 long.py:13(longest3)
     1000    0.000    0.000    0.031    0.000 long.py:22(<lambda>)
        1    0.000    0.000    0.000    0.000 timeit.py:143(setup)
        1    0.000    0.000    0.031    0.031 timeit.py:178(timeit)
        1    0.000    0.000    0.031    0.031 timeit.py:96(inner)
        1    0.000    0.000    0.000    0.000 {gc.disable}
        1    0.000    0.000    0.000    0.000 {gc.enable}
        1    0.000    0.000    0.000    0.000 {gc.isenabled}
        1    0.000    0.000    0.000    0.000 {globals}
     1000    0.000    0.000    0.000    0.000 {len}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
     1000    0.029    0.000    0.029    0.000 {sorted}
        2    0.000    0.000    0.000    0.000 {time.time}


None

Answer 2

def longest(strings): return max(strings, key=len)

您可以找到内置max函数here的文档。该密钥用于从每个字符串中提取值。

Answer 3

def longest_string(string_list):
    return max(len(s) for s in string_list)

示例运行：

l = ['abc', 'de', 'longest']
longest_string(l)
# 7

Answer 4

使用pandas vectorised str方法：

In [6]:

import pandas as pd
df = pd.DataFrame({'a':['asdsadasasdss','asdasdasdasasasdasdasd','asdsasdas']})
df

Out[6]:

                        a
0           asdsadasasdss
1  asdasdasdasasasdasdasd
2               asdsasdas

[3 rows x 1 columns]

In [8]:

df.a.str.len()

Out[8]:

0    13
1    22
2     9
Name: a, dtype: int64

In [9]:

df[df.a.str.len() ==max(df.a.str.len())]

Out[9]:

                        a
1  asdasdasdasasasdasdasd

[1 rows x 1 columns]

如何获取列表中最长字符串的长度（python）？

4 个答案: