正如Mark指出的，你应该使用.title():

"MyAwesomeString".title()

然而，如果你想让Django模板中的第一个字母大写，你可以这样做:

{{ "MyAwesomeString"|title }}

或者使用变量:

{{ myvar|title }}

如果访问[1:]，空字符串将引发错误。因此我会使用:

def my_uppercase(title):
    if not title:
       return ''
    return title[0].upper() + title[1:]

只大写第一个字母。

使用非均匀空格将字符串大写

我想补充一下@Amit Gupta关于非均匀空间的观点:

从最初的问题中，我们想要大写字符串s = 'the brown fox'中的每个单词。如果字符串s = 'the brown fox'有不均匀的空格。

def solve(s):
    # If you want to maintain the spaces in the string, s = 'the brown      fox'
    # Use s.split(' ') instead of s.split().
    # s.split() returns ['the', 'brown', 'fox']
    # while s.split(' ') returns ['the', 'brown', '', '', '', '', '', 'fox']
    capitalized_word_list = [word.capitalize() for word in s.split(' ')]
    return ' '.join(capitalized_word_list)

我非常喜欢这个答案:

复制粘贴版本的@jibberia回答:

def capitalize(line):
    return ' '.join([s[0].upper() + s[1:] for s in line.split(' ')])

但是我发送的一些行分离了一些空白的“字符，在尝试执行s[1:]时导致错误。可能有更好的方法，但我必须添加一个if len(s)>0，就像在

return ' '.join([s[0].upper() + s[1:] for s in line.split(' ') if len(s)>0])

如果str.title()对您不起作用，请自己大写。

将字符串拆分为单词列表 每个单词的第一个字母大写 把单词连接成一个字符串

一行程序:

>>> ' '.join([s[0].upper() + s[1:] for s in "they're bill's friends from the UK".split(' ')])
"They're Bill's Friends From The UK"

明显的例子:

input = "they're bill's friends from the UK"
words = input.split(' ')
capitalized_words = []
for word in words:
    title_case_word = word[0].upper() + word[1:]
    capitalized_words.append(title_case_word)
output = ' '.join(capitalized_words)

这里总结了不同的方法，以及一些需要注意的陷阱

它们将适用于所有这些输入:

""           => ""       
"a b c"      => "A B C"             
"foO baR"    => "FoO BaR"      
"foo    bar" => "Foo    Bar"   
"foo's bar"  => "Foo's Bar"    
"foo's1bar"  => "Foo's1bar"    
"foo 1bar"   => "Foo 1bar"     

Splitting the sentence into words and capitalizing the first letter then join it back together: # Be careful with multiple spaces, and empty strings # for empty words w[0] would cause an index error, # but with w[:1] we get an empty string as desired def cap_sentence(s): return ' '.join(w[:1].upper() + w[1:] for w in s.split(' ')) Without splitting the string, checking blank spaces to find the start of a word def cap_sentence(s): return ''.join( (c.upper() if i == 0 or s[i-1] == ' ' else c) for i, c in enumerate(s) ) Or using generators: # Iterate through each of the characters in the string # and capitalize the first char and any char after a blank space from itertools import chain def cap_sentence(s): return ''.join( (c.upper() if prev == ' ' else c) for c, prev in zip(s, chain(' ', s)) ) Using regular expressions, from steveha's answer: # match the beginning of the string or a space, followed by a non-space import re def cap_sentence(s): return re.sub("(^|\s)(\S)", lambda m: m.group(1) + m.group(2).upper(), s)

现在，这些是其他一些被发布的答案，如果我们将一个单词定义为句子的开头或空格后的任何东西，输入就不会像预期的那样工作:

.title () 返回s.title () #不需要的输出: "foO baR" => "foO baR" "foo's bar" => "foo's bar" "foo's1bar" => "foo's1bar" "foo 1bar" => "foo 1bar"

.capitalize()或.capwords() 返回' '.join(w.r esize () for s.split()中的w) #或 进口的字符串 返回string.capwords(年代) #不需要的输出: "foO baR" => "foO baR" "foo bar" => "foo bar" 使用' '作为分割将修复第二个输出，但不能修复第一个输出 返回' '.join(w.r esize () for w in s.s split(' ')) #或 进口的字符串 返回字符串。大写字符(s， ' ') #不需要的输出: "foO baR" => "foO baR"

.upper () 注意使用多个空格，这可以通过使用' '进行分割来修复(如答案顶部所示) 返回' ' . join (w [0] .upper () + w (1:) w s.split ()) #不需要的输出: "foo bar" => "foo bar"