blocks|key|1504025|text|def+is_ascii(s):
++++return+all(ord(c)+<+128+for+c+in+s)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1504026|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def is_ascii(s):
 return all(ord(c) &lt; 128 for c in s)
</code></pre>

blocks|key|1504003|text|我觉得你问的问题不对--|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504004|python中的字符串没有对应于'ascii‘、utf-8或任何其他编码的属性。字符串的来源(无论是从文件中读取，还是从键盘输入，等等)可能已经在ascii中编码了unicode字符串以生成您的字符串，但这是您需要找到答案的地方。|1504005|也许你可以问的问题是：“这个字符串是用ascii编码unicode字符串的结果吗？”--你可以通过尝试回答这个问题：|1504006|try:
++++mystring.decode('ascii')
except+UnicodeDecodeError:
++++print+"it+was+not+a+ascii-encoded+unicode+string"
else:
++++print+"It+may+have+been+an+ascii-encoded+unicode+string"|code-block|syntax|javascript|1504007|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think you are not asking the right question--

A string in python has no property corresponding to 'ascii', utf-8, or any other encoding. The source of your string (whether you read it from a file, input from a keyboard, etc.) may have encoded a unicode string in ascii to produce your string, but that's where you need to go for an answer.

Perhaps the question you can ask is: "Is this string the result of encoding a unicode string in ascii?" -- This you can answer
 by trying:

<pre><code>try:
 mystring.decode('ascii')
except UnicodeDecodeError:
 print "it was not a ascii-encoded unicode string"
else:
 print "It may have been an ascii-encoded unicode string"
</code></pre>

blocks|key|1504447|text|在Python+3中，我们可以将字符串编码为UTF-8，然后检查长度是否保持不变。如果是，则原始字符串为ASCII。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504448|def+isascii(s):
++++"""Check+if+the+characters+in+string+s+are+in+ASCII,+U%2B0-U%2B7F."""
++++return+len(s)+==+len(s.encode())|code-block|syntax|javascript|1504449|要进行检查，请传递测试字符串：|1504450|>>>+isascii("♥O◘♦♥O◘♦")
False
>>>+isascii("Python")
True|1504451|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

In Python 3, we can encode the string as UTF-8, then check whether the length stays the same. If so, then the original string is ASCII.
<pre class="lang-py prettyprint-override"><code>def isascii(s):
 &quot;&quot;&quot;Check if the characters in string s are in ASCII, U+0-U+7F.&quot;&quot;&quot;
 return len(s) == len(s.encode())
</code></pre>
To check, pass the test string:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; isascii(&quot;♥O◘♦♥O◘♦&quot;)
False
&gt;&gt;&gt; isascii(&quot;Python&quot;)
True
</code></pre>

blocks|key|1504860|text|Python3.7中的新特性(bpo32677)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1504861|不再繁琐/低效的ascii检查字符串，新的内置str/bytes/bytearray方法-+.isascii()将检查字符串是否为ascii。|style|CODE|1504862|print("is+this+ascii?".isascii())
#+True|code-block|syntax|javascript|1504863|entityMap|0|LINK|mutability|MUTABLE|url|https://bugs.python.org/issue32677|1|https://docs.python.org/3/library/stdtypes.html#str.isascii^0|F|8|0|0|N|3|R|5|X|9|1A|A|1A|A|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@$A|11|B|12|F|G]|$A|13|B|14|F|G]|$A|15|B|16|F|G]|$A|17|B|18|F|G]]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|H|3|I|5|J|7|1C|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h3>New in Python 3.7 (<a href="https://bugs.python.org/issue32677" rel="noreferrer">bpo32677</a>)</h3>

No more tiresome/inefficient ascii checks on strings, new built-in <code>str</code>/<code>bytes</code>/<code>bytearray</code> method - <a href="https://docs.python.org/3/library/stdtypes.html#str.isascii" rel="noreferrer"><code>.isascii()</code></a> will check if the strings is ascii.

<pre><code>print("is this ascii?".isascii())
# True
</code></pre>

blocks|key|1504650|text|Vincent+Marchetti的想法是正确的，但在Python3中，str.decode已被弃用。在Python3中，您可以使用str.encode进行相同的测试|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1504651|try:
++++mystring.encode('ascii')
except+UnicodeEncodeError:
++++pass++#+string+is+not+ascii
else:
++++pass++#+string+is+ascii|code-block|syntax|javascript|1504652|注意，您要捕获的异常也已从UnicodeDecodeError更改为UnicodeEncodeError。|1504653|entityMap^0|10|A|1U|A|0|0|D|I|Y|I|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Vincent Marchetti has the right idea, but <code>str.decode</code> has been deprecated in Python 3. In Python 3 you can make the same test with <code>str.encode</code>:

<pre><code>try:
 mystring.encode('ascii')
except UnicodeEncodeError:
 pass # string is not ascii
else:
 pass # string is ascii
</code></pre>

Note the exception you want to catch has also changed from <code>UnicodeDecodeError</code> to <code>UnicodeEncodeError</code>.

blocks|key|1504167|text|您的问题是不正确的；您看到的错误不是您如何构建python的结果，而是字节字符串和unicode字符串混淆的结果。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504168|字节字符串(例如，python语法中的"foo“或”bar“)是八位字节的序列；数字范围为0-255。Unicode字符串(例如u"foo“或u‘’bar‘)是unicode码位的序列；数字范围为0-1112064。但是您似乎对characteré感兴趣，它(在您的终端中)是一个表示单个字符的多字节序列。|1504169|而不是ord(u'é')，试试这个：|offset|length|style|CODE|1504170|>>>+[ord(x)+for+x+in+u'é']|code-block|syntax|javascript|1504171|它告诉你"é“代表的是哪一系列的代码点。它可能给你233，也可能给你101,770。|1504172|用unichr()来代替chr()来逆转这一点|1504173|>>>+unichr(233)
u'\xe9'|1504174|这个字符实际上可以用单个或多个unicode“代码点”表示，这些代码点本身表示字素或字符。它要么是"e带有一个尖锐的重音(即，代码点233)"，要么是"e“(代码点101)，后跟”前一个字符的一个尖锐的重音“(代码点770)。因此，可以将这个完全相同的字符表示为Python数据结构u'e\u0301'或u'\u00e9'。|1504175|大多数情况下，您不应该关心这一点，但如果您正在迭代unicode字符串，这可能会成为一个问题，因为迭代是按代码点而不是按可分解字符进行的。换句话说，就是len(u'e\u0301')+==+2和len(u'\u00e9')+==+1。如果这对您很重要，您可以使用unicodedata.normalize在合成表单和分解表单之间进行转换。|1504176|The+Unicode+Glossary可以通过指出每个特定术语如何引用文本表示的不同部分来帮助您理解其中的一些问题，这比许多程序员意识到的要复杂得多。|1504177|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/unicodedata.html#unicodedata.normalize|1|http://unicode.org/glossary/^0|0|0|3|9|0|0|0|1|8|C|5|0|0|3X|A|48|9|0|24|K|2P|J|3N|L|3N|L|0|0|0|K|1|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1C|8|@$F|1D|G|1E|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|1F|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1H|8|@$F|1I|G|1J|H|I]|$F|1K|G|1L|H|I]]|9|@]|A|$]]|$1|S|3|T|5|L|7|1M|8|@]|9|@]|A|$M|N]]|$1|U|3|V|5|6|7|1N|8|@$F|1O|G|1P|H|I]|$F|1Q|G|1R|H|I]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1S|8|@$F|1T|G|1U|H|I]|$F|1V|G|1W|H|I]|$F|1X|G|1Y|H|I]]|9|@$F|1Z|G|20|1|21]]|A|$]]|$1|Y|3|Z|5|6|7|22|8|@]|9|@$F|23|G|24|1|25]]|A|$]]|$1|10|3|-4|5|6|7|26|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]|18|$5|13|14|15|A|$16|19]]]]

Your question is incorrect; the error you see is not a result of how you built python, but of a confusion between byte strings and unicode strings.

Byte strings (e.g. "foo", or 'bar', in python syntax) are sequences of octets; numbers from 0-255. Unicode strings (e.g. u"foo" or u'bar') are sequences of unicode code points; numbers from 0-1112064. But you appear to be interested in the character é, which (in your terminal) is a multi-byte sequence that represents a single character.

Instead of <code>ord(u'é')</code>, try this:

<pre><code>&gt;&gt;&gt; [ord(x) for x in u'é']
</code></pre>

That tells you which sequence of code points "é" represents. It may give you [233], or it may give you [101, 770].

Instead of <code>chr()</code> to reverse this, there is <code>unichr()</code>:

<pre><code>&gt;&gt;&gt; unichr(233)
u'\xe9'
</code></pre>

This character may actually be represented either a single or multiple unicode "code points", which themselves represent either graphemes or characters. It's either "e with an acute accent (i.e., code point 233)", or "e" (code point 101), followed by "an acute accent on the previous character" (code point 770). So this exact same character may be presented as the Python data structure <code>u'e\u0301'</code> or <code>u'\u00e9'</code>.

Most of the time you shouldn't have to care about this, but it can become an issue if you are iterating over a unicode string, as iteration works by code point, not by decomposable character. In other words, <code>len(u'e\u0301') == 2</code> and <code>len(u'\u00e9') == 1</code>. If this matters to you, you can convert between composed and decomposed forms by using <a href="http://docs.python.org/library/unicodedata.html#unicodedata.normalize" rel="noreferrer"><code>unicodedata.normalize</code></a>.

<a href="http://unicode.org/glossary/" rel="noreferrer">The Unicode Glossary</a> can be a helpful guide to understanding some of these issues, by pointing how how each specific term refers to a different part of the representation of text, which is far more complicated than many programmers realize.

blocks|key|1719377|text|最近遇到了这样的事情--以供将来参考|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1719378|import+chardet

encoding+=+chardet.detect(string)
if+encoding['encoding']+==+'ascii':
++++print+'string+is+in+ascii'|code-block|syntax|javascript|1719379|您可以将其与以下内容一起使用：|1719380|string_ascii+=+string.decode(encoding['encoding']).encode('ascii')|1719381|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Ran into something like this recently - for future reference

<pre><code>import chardet

encoding = chardet.detect(string)
if encoding['encoding'] == 'ascii':
 print 'string is in ascii'
</code></pre>

which you could use with:

<pre><code>string_ascii = string.decode(encoding['encoding']).encode('ascii')
</code></pre>

blocks|key|1719162|text|这样做怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1719163|import+string

def+isAscii(s):
++++for+c+in+s:
++++++++if+c+not+in+string.ascii_letters:
++++++++++++return+False
++++return+True|code-block|syntax|javascript|1719164|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|$]]

How about doing this?

<pre><code>import string

def isAscii(s):
 for c in s:
 if c not in string.ascii_letters:
 return False
 return True
</code></pre>

blocks|key|1504545|text|要从Python2.6(以及Python3.x)改进Alexander的解决方案，您可以使用助手模块curses.ascii，并使用curses.ascii.isascii()函数或各种其他函数：https://docs.python.org/2.6/library/curses.ascii.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1504546|from+curses+import+ascii

def+isascii(s):
++++return+all(ascii.isascii(c)+for+c+in+s)|code-block|syntax|javascript|1504547|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2.6/library/curses.ascii.html^0|2Q|1H|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

To improve Alexander's solution from the Python 2.6 (and in Python 3.x) you can use helper module curses.ascii and use curses.ascii.isascii() function or various other: <a href="https://docs.python.org/2.6/library/curses.ascii.html" rel="nofollow">https://docs.python.org/2.6/library/curses.ascii.html</a>

<pre><code>from curses import ascii

def isascii(s):
 return all(ascii.isascii(c) for c in s)
</code></pre>

blocks|key|1719076|text|您可以使用接受Posix标准[:ASCII:]定义的正则表达式库。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1719077|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You could use the regular expression library which accepts the Posix standard [[:ASCII:]] definition.

blocks|key|1719888|text|与@RogerDahl的answer类似，但通过否定字符类并使用search而不是find_all或match更有效。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1719889|>>>+import+re
>>>+re.search('[%5E\x00-\x7F]',+'Did+you+catch+that+\x00?')+is+not+None
False
>>>+re.search('[%5E\x00-\x7F]',+'Did+you+catch+that+\xFF?')+is+not+None
True|code-block|syntax|javascript|1719890|我想正则表达式已经针对此进行了很好的优化。|1719891|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/32869248/623735^0|15|8|1E|5|C|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Like @RogerDahl's <a href="https://stackoverflow.com/a/32869248/623735">answer</a> but it's more efficient to short-circuit by negating the character class and using search instead of <code>find_all</code> or <code>match</code>.

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; re.search('[^\x00-\x7F]', 'Did you catch that \x00?') is not None
False
&gt;&gt;&gt; re.search('[^\x00-\x7F]', 'Did you catch that \xFF?') is not None
True
</code></pre>

I imagine a regular expression is well-optimized for this.

blocks|key|1719791|text|import+re

def+is_ascii(s):
++++return+bool(re.match(r'[\x00-\x7F]%2B$',+s))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1719792|要包含空字符串作为ASCII码，请将%2B更改为*。|unstyled|offset|length|style|CODE|1719793|entityMap^0|0|I|1|M|1|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>import re

def is_ascii(s):
 return bool(re.match(r'[\x00-\x7F]+$', s))
</code></pre>

To include an empty string as ASCII, change the <code>+</code> to <code>*</code>.

blocks|key|1504357|text|为了防止代码崩溃，您可能希望使用try-except来捕获TypeErrors|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1504358|>>>+ord("¶")
Traceback+(most+recent+call+last):
++File+"<stdin>",+line+1,+in+<module>
TypeError:+ord()+expected+a+character,+but+string+of+length+2+found|code-block|syntax|javascript|1504359|例如|1504360|def+is_ascii(s):
++++try:
++++++++return+all(ord(c)+<+128+for+c+in+s)
++++except+TypeError:
++++++++return+False|1504361|entityMap^0|G|A|T|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

To prevent your code from crashes, you maybe want to use a <code>try-except</code> to catch <code>TypeErrors</code>

<pre><code>&gt;&gt;&gt; ord("¶")
Traceback (most recent call last):
 File "&lt;stdin&gt;", line 1, in &lt;module&gt;
TypeError: ord() expected a character, but string of length 2 found
</code></pre>

For example 

<pre><code>def is_ascii(s):
 try:
 return all(ord(c) &lt; 128 for c in s)
 except TypeError:
 return False
</code></pre>

blocks|key|1719317|text|我使用以下内容来确定字符串是ascii还是unicode：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1719318|>>+print+'test+string'.__class__.__name__
str
>>>+print+u'test+string'.__class__.__name__
unicode
>>>+|code-block|syntax|javascript|1719319|然后只需使用条件块来定义函数：|1719320|def+is_ascii(input):
++++if+input.__class__.__name__+==+"str":
++++++++return+True
++++return+False|1719321|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I use the following to determine if the string is ascii or unicode:

<pre><code>&gt;&gt; print 'test string'.__class__.__name__
str
&gt;&gt;&gt; print u'test string'.__class__.__name__
unicode
&gt;&gt;&gt; 
</code></pre>

Then just use a conditional block to define the function:

<pre><code>def is_ascii(input):
 if input.__class__.__name__ == "str":
 return True
 return False
</code></pre>

I want to I check whether a string is in ASCII or not.

I am aware of <code>ord()</code>, however when I try <code>ord('é')</code>, I have <code>TypeError: ord() expected a character, but string of length 2 found</code>. I understood it is caused by the way I built Python (as explained in <a href="http://docs.python.org/library/functions.html#ord" rel="noreferrer"><code>ord()</code>'s documentation</a>). 

Is there another way to check?

How to check if a string in Python is in ASCII?

我想检查一个字符串是否为ASCII格式。我知道ord()，但是当我尝试ord('é')时，我使用的是TypeError: ord() expected a character, but string of length 2 found。我知道这是由我构建Python的方式引起的(正如中解释的那样)。有没有其他检查的方法？

问如何检查Python中的字符串是否为ASCII？
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问如何检查Python中的字符串是否为ASCII？
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