预期的字符串或缓冲区(在re.sub中)

时间:2014-02-07 04:16:01

标签: python

在SO之前看过类似的问题(herehere),我知道re.sub需要一个字符串(我相信,我提供的)但我不知道以下代码中的错误:

tuples = re.findall(r'id":"(.*?)".*?name":"(.*?)"', response.text, re.DOTALL)
outfile = open("badEXtsWithIDs.csv", "wb")
print "Writing into CSV"
writer = csv.writer(outfile)
for entry in tuples:
    writeName = re.sub(r'\W', " ", entry)
    writer.writerow(writeName)

我认为re.sub需要一个str变量,但不是str的入口吗?我在TypeError: expected string or buffer的行上收到错误:re.sub。任何帮助表示赞赏。

1 个答案:

答案 0 :(得分:5)

当您有多个匹配组时,re.findall会返回list个n - tuple s:

re.findall('(foo).(bar)', 'foo foo bar foo|bar')
Out[5]: [('foo', 'bar'), ('foo', 'bar')]

很清楚entry中的每个tuples都是tuple。当您将tuple传递给re.sub时,它会抱怨。

tuples = re.findall('(foo).(bar)', 'foo foo bar foo|bar')

for entry in tuples:
    re.sub('oo','ox',entry)

...
/usr/lib/python3.3/re.py in sub(pattern, repl, string, count, flags)
    168     a callable, it's passed the match object and must return
    169     a replacement string to be used."""
--> 170     return _compile(pattern, flags).sub(repl, string, count)
    171 
    172 def subn(pattern, repl, string, count=0, flags=0):

TypeError: expected string or buffer

所以,做点别的。也许使用map

for entry in tuples:
    print(' '.join(map(lambda s: re.sub('oo','ox',s),entry)))

fox bar
fox bar

或者更可读的理解

writer.writerow([re.sub(r'\W', " ",s) for s in entry])