随机大写字符串中的字母[重复]

Question

我想将字符串中的每个字母随机大写或小写。我是在 python 中使用字符串的新手，但我认为因为字符串是不可变的，所以我不能执行以下操作：

i =0             
for c in sentence:
    case = random.randint(0,1)
    print("case = ", case)
    if case == 0:
        print("here0")
        sentence[i] = sentence[i].lower()
    else:
        print("here1")
        sentence[i] = sentence[i].upper()
    i += 1
print ("new sentence = ", sentence)

并得到错误：TypeError: 'str' object does not support item assignment

但是我还能怎么做呢？

Answer 1

您可以将str.join 与这样的生成器表达式一起使用：

from random import choice
sentence = 'Hello World'
print(''.join(choice((str.upper, str.lower))(c) for c in sentence))

样本输出：

heLlo WORLd

Answer 2

构建一个新字符串。

这是一个对您的原始代码稍作改动的解决方案：

>>> import random
>>> 
>>> def randomcase(s):
...:    result = ''
...:    for c in s:
...:        case = random.randint(0, 1)
...:        if case == 0:
...:            result += c.upper()
...:        else:
...:            result += c.lower()
...:    return result
...:
...:
>>> randomcase('Hello Stackoverflow!')
>>> 'hElLo StaCkoVERFLow!'

edit: 删除了我的oneliners，因为我更喜欢blhsing。

Answer 3

import random
sentence='quick test'
print(''.join([char.lower() if random.randint(0,1) else char.upper() \
                   for char in sentence]))

快速测试

Answer 4

只需将字符串实现更改为列表实现即可。由于字符串是不可变的，因此您无法更改对象内部的值。但是Lists 可以，所以我只更改了您的那部分代码。请注意，有更好的方法可以做到这一点，关注here

import random
sentence = "This is a test sentence" # Strings are immutable
i =0
new_sentence = [] # Lists are mutable sequences
for c in sentence:
    case = random.randint(0,1)
    print("case = ", case)
    if case == 0:
        print("here0")
        new_sentence += sentence[i].lower() # append to the list
    else:
        print("here1")
        new_sentence += sentence[i].upper() # append to the list
    i += 1
print ("new sentence = ", new_sentence)

# to print as string
new_sent = ''.join(new_sentence)
print(new_sent)

Answer 5

你可以像下面那样做

char_list = []            
for c in sentence:
    ucase = random.randint(0,1)
    print("case = ", case)
    if ucase:
        print("here1")
        char_list.append(c.upper())
    else:
        print("here0")
        char_list.append(c.lower())
print ("new sentence = ", ''.join(char_list))

Answer 6

不涉及 python 循环的一种方法是将其发送到 numpy 并对其进行矢量化操作。例如：

import numpy as np
def randomCapitalize(s):
    s  = np.array(s, 'c').view('u1')
    t  = np.random.randint(0, 2, len(s), 'u1') # Temporary array
    t *= s != 32 # ASCII 32 (i.e. space) should not be lowercased
    t *= 32 # Decrease ASCII by 32 to lowercase
    s -= t
    return s.view('S' + str(len(s)))[0]
randomCapitalize('hello world jfwojeo jaiofjowejfefjawivj a jofjawoefj')

哪个输出：

b'HELLO WoRlD jFwoJEO JAioFjOWeJfEfJAWIvJ A JofjaWOefj'

这个解决方案应该相当快，尤其是对于长字符串。这种方法有两个限制：

• 输入必须完全小写。你可以先试试.lower()，但这在技术上效率很低。

• 需要特别注意非 a-to-z 字符。在上面的例子中，只处理了空格

您可以通过替换同时处理更多的特殊字符

t *= s != 32

与

# using space, enter, comma, period as example
t *= np.isin(s, list(map(ord, ' \n,.')), invert=True)

例如：

s = 'ascii table and description. ascii stands for american standard code for information interchange. computers can only understand numbers, so an ascii code is the numerical representation of a character such as'
randomCapitalize(s)

哪个输出：

b'ascII tABLe AnD descRiptIOn. ascii sTaNds for AmEricAN stanDaRD codE FOr InForMAtION iNTeRCHaNge. ComPUtERS can onLY UNdersTand nUMBers, So An asCIi COdE IS tHE nuMERIcaL rEPrEsEnTATion Of a CHaractEr such as'