分析
数字中经常是3个数字一组，之后跟一个逗号，因此规律为：***，***，***
正则式[a-z]+,[a-z]?
import re
sen = "abc,123,456,789,mnp"
p = pile("d+,d+?")
for com in p.finditer(sen):
mm = com.group()
print "hi:", mm
print "sen_before:", sen
sen = sen.replace(mm, mm.replace(",", ""))
print "sen_back:", sen, 'n'
技巧
使用函数finditer(string[, pos[, endpos]]) | re.finditer(pattern, string[, flags]):
搜索string，返回一个顺序访问每一个匹配结果(Match对象)的迭代器。
sen = "abc,123,456,789,mnp"
while 1:
mm = re.search("d,d", sen)
if mm:
mm = mm.group()
sen = sen.replace(mm, mm.replace(",", ""))
print sen
else:
break
这样的程序针对具体问题，即数字3位一组，如果数字混杂与字母间，干掉数字间的逗号，即把“abc,123,4,789,mnp”转化为“abc,1234789,mnp”
更具体的是找正则式“数字，数字”找到后用去掉逗号的替换
sen = "abc,123,4,789,mnp"
while 1:
mm = re.search("d,d", sen)
if mm:
mm = mm.group()
sen = sen.replace(mm, mm.replace(",", ""))
print sen
else:
break
print sen
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