blocks|key|3760804|text|你可以像这样在你的函数中使用generator：-|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3760805|def+isInAlphabeticalOrder(word):
++++return+all((word[i%2B1]+>=+word[i]+for+i+in+range(len(word)+-+1)))|code-block|syntax|javascript|3760806|生成器从给定范围中获取i的每个值，并将该索引处的字符与前一个索引处的字符进行比较。并将所有比较结果传递给all函数，如果所有值都为True，则该函数将返回True。|3760807|>>>+def+isInAlphabeticalOrder(word):
++++++++return+all((word[i%2B1]+>=+word[i]+for+i+in+range(len(word)+-+1)))

>>>+isInAlphabeticalOrder("rohit")
False
>>>+isInAlphabeticalOrder("aabc")
True
>>>+isInAlphabeticalOrder("abc")
True|3760808|3760809|当然，这并不考虑case-insensitivity。如果需要考虑，则将返回语句更改为：-|3760810|return+all((str.lower(word[i%2B1])+>=+str.lower(word[i])+for+i+in+range(len(word)+-+1)))|3760811|entityMap^0|E|9|0|0|B|1|1G|3|1T|4|25|4|0|0|0|8|I|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|T|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|U|$]]

You can use <code>generator</code> in your function like this: -

<pre><code>def isInAlphabeticalOrder(word):
 return all((word[i+1] &gt;= word[i] for i in range(len(word) - 1)))
</code></pre>

The generator gets each value of <code>i</code> from the given range, and compare the character at that index with the one in the previous index. And all the comparison results are passed to <code>all</code> function, which will return <code>True</code> if all the values are <code>True</code>.

<pre><code>&gt;&gt;&gt; def isInAlphabeticalOrder(word):
 return all((word[i+1] &gt;= word[i] for i in range(len(word) - 1)))

&gt;&gt;&gt; isInAlphabeticalOrder("rohit")
False
&gt;&gt;&gt; isInAlphabeticalOrder("aabc")
True
&gt;&gt;&gt; isInAlphabeticalOrder("abc")
True
</code></pre>

<hr>

Of course that does not consider <code>case-insensitivity</code>. If you want to consider it, then change the return statement to: -

<pre><code>return all((str.lower(word[i+1]) &gt;= str.lower(word[i]) for i in range(len(word) - 1)))
</code></pre>

blocks|key|4241346|text|这是一种简单的(和Python惯用的)方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4241347|def+isInAlphabeticalOrder(word):
++++return+word==''.join(sorted(word))

>>>+isInAlphabeticalOrder('abc')
True
>>>+isInAlphabeticalOrder('acb')++++
False|code-block|syntax|javascript|4241348|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is a simple (and Python idiomatic) way to do this:

<pre><code>def isInAlphabeticalOrder(word):
 return word==''.join(sorted(word))

&gt;&gt;&gt; isInAlphabeticalOrder('abc')
True
&gt;&gt;&gt; isInAlphabeticalOrder('acb') 
False
</code></pre>

blocks|key|411538|text|这样做的优点是O(n)+(对字符串进行排序是O(n+log+n))。在Python中，如果一个字符(或字符串)按字母顺序出现在它之前，那么它就是“小于”另一个字符，所以为了查看一个字符串是否按字母顺序排列，我们只需要比较每一对相邻的字符。还要注意，您需要使用range(len(word)+-+1)而不是range(len(word))，否则您将在循环的最后一次迭代中超出字符串的界限。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|411539|def+isInAlphabeticalOrder(word):
++++for+i+in+range(len(word)+-+1):
++++++++if+word[i]+>+word[i+%2B+1]:
++++++++++++return+False
++++return+True|code-block|syntax|javascript|411540|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This has the advantage of being O(n) (sorting a string is O(n log n)). A character (or string) in Python is "less than" another character if it comes before it in alphabetical order, so in order to see if a string is in alphabetical order we just need to compare each pair of adjacent characters. Also, note that you take range(len(word) - 1) instead of range(len(word)) because otherwise you will overstep the bounds of the string on the last iteration of the loop.

<pre><code>def isInAlphabeticalOrder(word):
 for i in range(len(word) - 1):
 if word[i] &gt; word[i + 1]:
 return False
 return True
</code></pre>

blocks|key|4241497|text|试着这样做，作为一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4241498|all(x+<=+y+for+x,+y+in+zip(word,+word[1:]))|code-block|syntax|javascript|4241499|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this, as a one-liner:

<pre><code>all(x &lt;= y for x, y in zip(word, word[1:]))
</code></pre>

blocks|key|5970274|text|如果单词按字母顺序排列，程序将返回true，否则返回false。第二个参数wordList被初始化为None，这意味着程序将检查您输入的任何单词，包括数字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5970275|def+isAlphabeticalOrder(word,+wordList+=+None):
if+(len(word)+>+0):
++++curr+=+word[0]
for+letter+in+word:
++++if+(curr+>+letter):
++++++++return+False
++++else:
++++++++curr+=+letter
if+wordList+is+None:
++++return+True
return+word+in+wordList|code-block|syntax|javascript|5970276|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The program will return true if the word is alphabetically arranged or false otherwise. The second argument wordList is initialized to None meaning the program will check for any word that you put in including numbers:

<pre><code>def isAlphabeticalOrder(word, wordList = None):
if (len(word) &gt; 0):
 curr = word[0]
for letter in word:
 if (curr &gt; letter):
 return False
 else:
 curr = letter
if wordList is None:
 return True
return word in wordList
</code></pre>

blocks|key|5970314|text|有几个答案已经解决了实际的字符串比较。但我想补充一点关于你的返回逻辑。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5970315|对于初学者来说，通常会编写如下代码：|5970316|if+something+==+somethingElse:
++return+True
else:
++return+False|code-block|syntax|javascript|5970317|代码总是可以像这样简化：|5970318|return+something+==+somethingElse|5970319|如果该代码一开始没有意义，那么它就会读作“将某些内容与SomethingElse进行比较，并返回比较结果”。|5970320|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Several answers have already addressed the actual string comparison. But I want to add a bit about your return logic.

It is common for beginners to write code like:

<pre><code>if something == somethingElse:
 return True
else:
 return False
</code></pre>

That code can always be simplified like this:

<pre><code>return something == somethingElse
</code></pre>

If that code doesn't make sense at first it reads as, "Compare something to SomethingElse, and return the result of the comparison".

blocks|key|4241592|text|这是最简单的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4241593|def+alphabetical(word):
++++return+list(word)+==+sorted(word)|code-block|syntax|javascript|4241594|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is the easiest:

<pre><code>def alphabetical(word):
 return list(word) == sorted(word)
</code></pre>

I've put together the following code to check if a string/word is alphabetically ordered:

<pre><code>def isInAlphabeticalOrder(word):
 word1=sorted(word)
 word2=[]
 for i in word:
 word2.append(i)
 if word2 == word1:
 return True
 else:
 return False
</code></pre>

but I feel like there must be a more efficient way (fewer lines of code) to check other than turning the strings into lists. Isn't there a operand to sort strings alphabetically without turning each char into a list? Can anyone suggest a more efficient way?

checking if a string is in alphabetical order in python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我将以下代码放在一起检查字符串/单词是否按字母顺序排序：def isInAlphabeticalOrder(word):    word1=sorted(word)    word2=[]    for i in word:        word2.append(i)    if word2 == word1:   ...

问在python中检查字符串是否按字母顺序排列
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问在python中检查字符串是否按字母顺序排列EN