blocks|key|1400495|text|line[5:]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1400496|给出前五个字符之后的字符。|unstyled|1400497|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>line[5:]
</code></pre>

gives you characters after the first five.

blocks|key|1400544|text|line[5:]会给出你想要的子串。搜索introduction并查找“slice+notation”|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1400545|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/introduction.html^0|0|8|K|C|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<code>line[5:]</code> will give the substring you want. Search the <a href="http://docs.python.org/tutorial/introduction.html" rel="nofollow noreferrer">introduction</a> and look for 'slice notation'

blocks|key|1615297|text|如果字符串是固定的，您可以简单地使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1615298|if+line.startswith("Path="):
++++return+line[5:]|code-block|syntax|javascript|1615299|它提供了从字符串中位置5开始的所有内容(字符串也是一个序列，因此这些序列操作符也可以在这里使用)。|1615300|或者，您可以在第一个=处拆分该行|offset|length|style|CODE|1615301|if+"="+in+line:
++++param,+value+=+line.split("=",1)|1615302|那么param就是"Path“，value就是第一个=之后的剩余部分。|1615303|entityMap^0|0|0|0|A|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

If the string is fixed you can simply use:

<pre><code>if line.startswith("Path="):
 return line[5:]
</code></pre>

which gives you everything from position 5 on in the string (a string is also a sequence so these sequence operators work here, too).

Or you can split the line at the first <code>=</code>:

<pre><code>if "=" in line:
 param, value = line.split("=",1)
</code></pre>

Then param is "Path" and value is the rest after the first =.

blocks|key|1400604|text|>>>+import+re

>>>+p+=+re.compile(r'path=(.*)',+re.IGNORECASE)

>>>+path+=+"path=c:\path"

>>>+re.match(p,+path).group(1)
'c:\\path'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1400605|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>&gt;&gt;&gt; import re

&gt;&gt;&gt; p = re.compile(r'path=(.*)', re.IGNORECASE)

&gt;&gt;&gt; path = "path=c:\path"

&gt;&gt;&gt; re.match(p, path).group(1)
'c:\\path'
</code></pre>

blocks|key|1615337|text|如果你知道列表的理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1615338|lines+=+[line[5:]+for+line+in+file.readlines()+if+line[:5]+==+"Path="]|code-block|syntax|javascript|1615339|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|$]]

If you know list comprehensions:

<pre><code>lines = [line[5:] for line in file.readlines() if line[:5] == "Path="]
</code></pre>

blocks|key|1400669|text|与索引[-1]相比，我更喜欢pop|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1400670|value+=+line.split("Path=",+1).pop()|code-block|syntax|javascript|1400671|至|1400672|value+=+line.split("Path=",+1)[1]
param,+value+=+line.split("Path=",+1)|1400673|entityMap^0|3|4|E|3|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|$]]

I prefer <code>pop</code> to indexing <code>[-1]</code>:

<pre><code>value = line.split("Path=", 1).pop()
</code></pre>

to

<pre><code>value = line.split("Path=", 1)[1]
param, value = line.split("Path=", 1)
</code></pre>

blocks|key|1615390|text|任何Python版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1615391|def+remove_prefix(text,+prefix):
++++return+text[len(prefix):]+if+text.startswith(prefix)+else+text|code-block|syntax|javascript|1615392|Python+3.9%2B|1615393|text.removeprefix(prefix)|1615394|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|$]]

Any Python version:
<pre><code>def remove_prefix(text, prefix):
 return text[len(prefix):] if text.startswith(prefix) else text
</code></pre>
Python 3.9+
<pre><code>text.removeprefix(prefix)
</code></pre>

blocks|key|1615426|text|为什么不呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1615427|if+line.startswith(prefix):
++++return+line.replace(prefix,+'',+1)|code-block|syntax|javascript|1615428|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|$]]

Or why not

<pre><code>if line.startswith(prefix):
 return line.replace(prefix, '', 1)
</code></pre>

blocks|key|1400737|text|这样怎么样..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1400738|>>>+line+=+r'path=c:\path'
>>>+line.partition('path=')
('',+'path=',+'c:\\path')|code-block|syntax|javascript|1400739|这个三元组就是head,+separator,+and+tail。|offset|length|1400740|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.7/library/stdtypes.html#str.partition^0|0|0|7|P|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

How about..

<pre><code>&gt;&gt;&gt; line = r'path=c:\path'
&gt;&gt;&gt; line.partition('path=')
('', 'path=', 'c:\\path')
</code></pre>

This triplet is the <a href="https://docs.python.org/3.7/library/stdtypes.html#str.partition" rel="nofollow noreferrer">head, separator, and tail</a>.

blocks|key|1400764|text|我猜这就是你要找的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1400765|++++def+findPath(i_file)+:
++++++++lines+=+open(+i_file+).readlines()
++++++++for+line+in+lines+:
++++++++++++if+line.startswith(+"Path="+):
++++++++++++++++output_line=line[(line.find("Path=")%2Blen("Path=")):]
++++++++++++++++return+output_line|code-block|syntax|javascript|1400766|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|$]]

I guess this what you are exactly looking for

<pre><code> def findPath(i_file) :
 lines = open( i_file ).readlines()
 for line in lines :
 if line.startswith( "Path=" ):
 output_line=line[(line.find("Path=")+len("Path=")):]
 return output_line
</code></pre>

blocks|key|1400793|text|流行音乐的版本不太正确。我认为你想要：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1400794|>>>+print('foofoobar'.split('foo',+1).pop())
foobar|code-block|syntax|javascript|1400795|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 pop version wasn't quite right. I think you want:

<pre><code>&gt;&gt;&gt; print('foofoobar'.split('foo', 1).pop())
foobar
</code></pre>

blocks|key|1400826|text|这里没有提到的另一个简单的一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1400827|value+=+line.split("Path=",+1)[-1]|code-block|syntax|javascript|1400828|这也适用于各种边缘情况：|1400829|>>>+print("prefixfoobar".split("foo",+1)[-1])
"bar"

>>>+print("foofoobar".split("foo",+1)[-1])
"foobar"

>>>+print("foobar".split("foo",+1)[-1])
"bar"

>>>+print("bar".split("foo",+1)[-1])
"bar"

>>>+print("".split("foo",+1)[-1])
""|1400830|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|$]]

Another simple one-liner that hasn't been mentioned here:

<pre><code>value = line.split("Path=", 1)[-1]
</code></pre>

This will also work properly for various edge cases:

<pre><code>&gt;&gt;&gt; print("prefixfoobar".split("foo", 1)[-1])
"bar"

&gt;&gt;&gt; print("foofoobar".split("foo", 1)[-1])
"foobar"

&gt;&gt;&gt; print("foobar".split("foo", 1)[-1])
"bar"

&gt;&gt;&gt; print("bar".split("foo", 1)[-1])
"bar"

&gt;&gt;&gt; print("".split("foo", 1)[-1])
""
</code></pre>

blocks|key|1400873|text|在没有编写函数的情况下，这将根据列表进行拆分，在本例中是‘先生%7C博士%7C夫人’，用1选择拆分后的所有内容，然后再次拆分并抓取任何元素。在下面的例子中，返回'Morris‘。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1400874|re.split('Mr.%7CDr.%7CMrs.',+'Mr.+Morgan+Morris')[1].split()[1]|code-block|syntax|javascript|1400875|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|$]]

without having a to write a function, this will split according to list, in this case 'Mr.|Dr.|Mrs.', select everything after split with [1], then split again and grab whatever element. In the case below, 'Morris' is returned.

<pre><code>re.split('Mr.|Dr.|Mrs.', 'Mr. Morgan Morris')[1].split()[1]
</code></pre>

blocks|key|1400941|text|为什么不使用带转义的正则表达式呢？%5E匹配一行的起始部分，而re.MULTILINE匹配每行。re.escape确保匹配是精确的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1400942|>>>+print(re.sub('%5E'+%2B+re.escape('path='),+repl='',+string='path=c:\path\nd:\path2',+flags=re.MULTILINE))
c:\path
d:\path2|code-block|syntax|javascript|1400943|entityMap^0|H|1|T|C|1A|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$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]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Why not using regex with escape?
<code>^</code> matches the initial part of a line and <code>re.MULTILINE</code> matches on each line. <code>re.escape</code> ensures that the matching is exact.

<pre><code>&gt;&gt;&gt; print(re.sub('^' + re.escape('path='), repl='', string='path=c:\path\nd:\path2', flags=re.MULTILINE))
c:\path
d:\path2
</code></pre>

blocks|key|1400983|text|这在技术上与其他答案非常相似，但没有重复的字符串操作，能够判断前缀是否存在，并且仍然具有很好的可读性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1400984|parts+=+the_string.split(prefix_to_remove,+1):
++++if+len(parts)+==+2:
++++++++#++do+things+with+parts[1]
++++++++pass|code-block|syntax|javascript|1400985|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 very similar in technique to other answers, but with no repeated string operations, ability to tell if the prefix was there or not, and still quite readable:

<pre><code>parts = the_string.split(prefix_to_remove, 1):
 if len(parts) == 2:
 # do things with parts[1]
 pass
</code></pre>

blocks|key|1401020|text|尝试以下代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1401021|if+line.startswith("Path="):+return+line[5:]|code-block|syntax|javascript|1401022|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 Following code 

<pre><code>if line.startswith("Path="): return line[5:]
</code></pre>

blocks|key|1615811|text|从Python+3.9开始，您可以使用removeprefix|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1615812|'Path=helloworld'.removeprefix('Path=')
#+'helloworld'|code-block|syntax|javascript|1615813|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.removeprefix^0|1|A|J|C|J|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Starting in <code>Python 3.9</code>, you can use <a href="https://docs.python.org/3/library/stdtypes.html#str.removeprefix" rel="noreferrer"><code>removeprefix</code></a>:
<pre><code>'Path=helloworld'.removeprefix('Path=')
# 'helloworld'
</code></pre>

blocks|key|1401085|text|可以尝试下面的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1401086|def+remove_suffix(string1,+suffix):
++++length+=+len(suffix)

++++if+string1[0:length]+==+suffix:
++++++++return+string1[length:]
++++else:
++++++++return+string1

suffix+=+"hello"
string1+=+"hello+world"

final_string+=+remove_suffix(string1,+suffix)
print+(final_string)|code-block|syntax|javascript|1401087|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 below method can be tried.
<pre><code>def remove_suffix(string1, suffix):
 length = len(suffix)

 if string1[0:length] == suffix:
 return string1[length:]
 else:
 return string1

suffix = &quot;hello&quot;
string1 = &quot;hello world&quot;

final_string = remove_suffix(string1, suffix)
print (final_string)
</code></pre>

I have some simple python code that searches files for a string e.g. <code>path=c:\path</code>, where the <code>c:\path</code> part may vary. The current code is:

<pre><code>def find_path(i_file):
 lines = open(i_file).readlines()
 for line in lines:
 if line.startswith("Path="):
 return # what to do here in order to get line content after "Path=" ?
</code></pre>

What is a simple way to get the text after <code>Path=</code>?

How to remove the left part of a string?

Python

我有一些简单的python代码来搜索文件中的字符串，例如path=c:\path，其中的c:\path部分可能会有所不同。当前代码为：def find_path(i_file):    lines = open(i_file).readlines()    for line in lines:        if li...

问如何删除字符串的左边部分？
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问如何删除字符串的左边部分？
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