如何findall()对 pandas 数据框的正则表达式序列? [英] How to findall() a sequence of regular expressions to a pandas dataframe?

问题描述

我正在使用熊猫 findall 函数.但是，我有几个正则表达式.这样，如何用熊猫findall N正则表达式?.

I am extracting some patterns with pandas findall function. However, I have several regular expressions. This, how can I findall N regular expressions with pandas?.

例如，假设我要提取特定列中的所有数字和所有日期:

For example, lets say that I would like to extract the all the numbers and all the dates inside an specific column:

在:

dfs = pd.DataFrame(data={'c1': ['This dataset 11/12/98 contains 5,000 rows, which were sampled from a 500,000 11/12/12 row dataset spanning the same time period. Throughout these analyses', 

                                'the number of events you count will be about 100 times smaller than they 11/12/78 actually were, but the 01/12/11 proportions of events will still generally be reflective that larger dataset. In this case, a sample is fine because our purpose is to learn methods of data analysis with Python, not to create 100% accurate recommendations to Watsi.']})

dfs

出局:

    c1
0   This dataset 11/12/98 contains 5,000 rows, whi...
1   the number of events you count will be about 1...

我试图这样做，但是出现以下错误:

I tried to, but I am getting the following error:

在:

dfs['patterns'] = dfs['c1'].str.findall([r'\d+',r'(\d+/\d+/\d+)']).apply(', '.join)

dfs

出局:

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-64-af2969e06a61> in <module>()
----> 1 dfs['patterns'] = dfs['c1'].str.findall([r'\d+',r'(\d+/\d+/\d+)']).apply(', '.join)
      2 dfs

/usr/local/lib/python3.5/site-packages/pandas/core/strings.py in wrapper2(self, pat, flags, **kwargs)
   1268 
   1269     def wrapper2(self, pat, flags=0, **kwargs):
-> 1270         result = f(self._data, pat, flags=flags, **kwargs)
   1271         return self._wrap_result(result)
   1272 

/usr/local/lib/python3.5/site-packages/pandas/core/strings.py in str_findall(arr, pat, flags)
    827     extractall : returns DataFrame with one column per capture group
    828     """
--> 829     regex = re.compile(pat, flags=flags)
    830     return _na_map(regex.findall, arr)
    831 

/usr/local/Cellar/python3/3.5.2_2/Frameworks/Python.framework/Versions/3.5/lib/python3.5/re.py in compile(pattern, flags)
    222 def compile(pattern, flags=0):
    223     "Compile a regular expression pattern, returning a pattern object."
--> 224     return _compile(pattern, flags)
    225 
    226 def purge():

/usr/local/Cellar/python3/3.5.2_2/Frameworks/Python.framework/Versions/3.5/lib/python3.5/re.py in _compile(pattern, flags)
    279     # internal: compile pattern
    280     try:
--> 281         p, loc = _cache[type(pattern), pattern, flags]
    282         if loc is None or loc == _locale.setlocale(_locale.LC_CTYPE):
    283             return p

TypeError: unhashable type: 'list'

因此，如何使用findall函数堆叠"，嵌套"或应用多个正则表达式? 我期望作为输出的是每个正则表达式的分辨率，在单列中用,分隔:

Therefore how can I "stack", "nest" or apply several regex with findall function?. What I expect as an output is the resolution of each regular expression separated by , in a single column:

   col
0  '11/12/98', '5', '000', '500', '000', '11/12/12'
1  '100', '11/12/78', '01/12/11', '100'

更新

我试图:

dfs['patterns'] = dfs['c1'].str.map(findall(),[r'\d+',r'(\d+/\d+/\d+)']).apply(', '.join)
dfs

推荐答案

仍然无法清除所需的输出. 但是请检查以下代码.

Still not clear your desired output. But please check below code.

dfs['patterns'] = dfs['c1'].str.findall(r'\d+\/\d+\/\d+|\d+')
print dfs['patterns'].sum()

['11/12/98', '5', '000', '500', '000', '11/12/12', '100', '11/12/78', '01/12/11', '100']

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