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python replace backslashes to slashes

开发者 https://www.devze.com 2023-03-11 23:01 出处:网络
How can I escape the backslashes in the str开发者_C百科ing: \'pictures\\12761_1.jpg\'? I know about raw string. How can I convert str to raw if I take \'pictures\\12761_1.jpg\' value from xml file fo

How can I escape the backslashes in the str开发者_C百科ing: 'pictures\12761_1.jpg'?

I know about raw string. How can I convert str to raw if I take 'pictures\12761_1.jpg' value from xml file for example?


You can use the string .replace() method along with rawstring.

Python 2:

>>> print r'pictures\12761_1.jpg'.replace("\\", "/")
pictures/12761_1.jpg

Python 3:

>>> print(r'pictures\12761_1.jpg'.replace("\\", "/"))
pictures/12761_1.jpg

There are two things to notice here:

  • Firstly to read the text as a drawstring by putting r before the string. If you don't give that, there will be a Unicode error here.
  • And also that there were two backslashes given inside the replace method's first argument. The reason for that is that backslash is a literal used with other letters to work as an escape sequence. Now you might wonder what is an escape sequence. So an escape sequence is a sequence of characters that doesn't represent itself when used inside string literal or character. It is composed of two or more characters starting with a backslash. Like '\n' represents a newline and similarly there are many. So to escape backslash itself which is usually an initiation of an escape sequence, we use another backslash to escape it.

I know the second part is bit confusing but I hope it made some sense.


You can also use split/join:

print "/".join(r'pictures\12761_1.jpg'.split("\\"))

EDITED:

The other way you may use is to prepare data during it's retrieving(e.g. the idea is to update string before assign to variable) - for example:

f = open('c:\\tst.txt', "r")
print f.readline().replace('\\','/')

>>>'pictures/12761_1.jpg\n'


I know it is not what you asked exactly, but I think this will work better. Tit's better to just have the names of your directories and use os.path.join(directory,filename)

"os.path.join(path, *paths) Join one or more path components intelligently. The return value is the concatenation of path and any members of *paths with exactly one directory separator (os.sep) following each non-empty part except the last, meaning that the result will only end in a separator if the last part is empty. If a component is an absolute path, all previous components are thrown away and joining continues from the absolute path component"

https://docs.python.org/2/library/os.path.html

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