re.sub(pattern, repl, string, count=0, flags=0)
Return the string obtained by replacing the leftmost non-overlapping occurrences of pattern in string by the replacement repl. If the pattern isn’t found, string is returned unchanged. repl can be a string or a function; if it is a string, any backslash escapes in
 
 it are processed. That is, \n is converted to a single newline character, \r is converted to a carriage return, and so forth. Unknown escapes such as \j are left alone. Backreferences, such as \6, are replaced with the substring matched by group 6 in the pattern. For example:

>>> re.sub(r'def\s+([a-zA-Z_][a-zA-Z_0-9]*)\s*\(\s*\):'
 
,
...        r'static PyObject*\npy_\1(void)\n{',
...        'def myfunc():')
'static PyObject*\npy_myfunc(void)\n{'
If repl is a function, it is called for every non-overlapping occurrence of pattern. The function takes a single match object argument, and returns the replacement string. For example:
 


>>> def dashrepl(matchobj):
...     if matchobj.group(0) == '-': return ' '
...     else: return '-'
>>> re.sub('-{1,2}', dashrepl, 'pro----gram-files')
'pro--gram files'
>>> re.sub(r'\sAND\s', ' & ', 'Baked Beans And Spam', flags=re.IGNORECASE)
'Baked Beans & Spam'
The pattern may be a string or an RE object.

The optional argument count is the maximum number of pattern occurrences to be replaced; countmust be a non-negative integer. If omitted or zero, all occurrences will be replaced. Empty matches for the pattern are replaced only when not adjacent to a previous match, so sub('x*', '-', 'abc') returns '-a-b-c-'.

In addition to character escapes and backreferences as described above, \g<name> will use the substring matched by the group named name, as defined by the (?P<name>...) syntax. \g<number>uses the corresponding group number; \g<2> is therefore equivalent to \2, but isn’t ambiguous in a replacement such as \g<2>0. \20 would be interpreted as a reference to group 20, not a reference to group 2 followed by the literal character '0'. The backreference \g<0> substitutes in the entire substring matched by the RE.

Changed in version 2.7: Added the optional flags argument.

簡單來說，pattern 就是一個需要被替換的正則表示式，當其匹配成功後，就會用 repl 進行替換，而 string 就是需要被替換的字串。

比如，我們需要對一個字串進行處理，首先需要刪除括號內所有內容，包括括號，其次，刪除空格，然後按照逗號將其進行分割

s = '物流 企業, 生產效率, 資料包絡分析(DEA),Window Analysis,'
r1 = re.compile('[(（].*?[)）]|\s|,$')
# [(（] 匹配英文(  或者中文（
# [(（]\w*?[)）] 匹配以中文括號或者英文括號括起來的 \w*?
# \w*? 匹配字母、數字、下劃線，重複任意次，儘可能少重複
# | 邏輯或
# \s 任意空白符
# ,$ 從最後開始的一個逗號
s1 = re.sub(r1,'',s)   # 對於正則表示式進行 '' 替換，等效於刪除正則表示式可以匹配的內容
# 按照逗號分割
s2 = s1.split(',')   # 按照逗號將其分割
for x in s2:
    print x

s1 的結果為 物流企業,生產效率,資料包絡分析,WindowAnalysis