Python 计算熊猫数据框中某些单词的出现次数
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Count occurrences of certain words in pandas dataframe
提问by Nilani Algiriyage
I want to count number of occurrences of certain words in a data frame. I know using "str.contains"
我想计算数据框中某些单词的出现次数。我知道使用“str.contains”
a = df2[df2['col1'].str.contains("sample")].groupby('col2').size()
n = a.apply(lambda x: 1).sum()
Currently I'm using the above code. Is there a method to match regular expression and get the count of occurrences? In my case I have a large dataframe and I want to match around 100 strings.
目前我正在使用上面的代码。是否有匹配正则表达式并获取出现次数的方法？就我而言，我有一个大数据框，我想匹配大约 100 个字符串。
采纳答案by Andy Hayden
Update: Original answer counts those rows which contain a substring.
更新：原始答案计算那些包含子字符串的行。
To count all the occurrences of a substring you can use .str.count:
要计算子字符串的所有出现次数，您可以使用.str.count：
In [21]: df = pd.DataFrame(['hello', 'world', 'hehe'], columns=['words'])
In [22]: df.words.str.count("he|wo")
Out[22]:
0 1
1 1
2 2
Name: words, dtype: int64
In [23]: df.words.str.count("he|wo").sum()
Out[23]: 4
The str.containsmethod accepts a regular expression:
该str.contains方法接受一个正则表达式：
Definition: df.words.str.contains(self, pat, case=True, flags=0, na=nan)
Docstring:
Check whether given pattern is contained in each string in the array
Parameters
----------
pat : string
Character sequence or regular expression
case : boolean, default True
If True, case sensitive
flags : int, default 0 (no flags)
re module flags, e.g. re.IGNORECASE
na : default NaN, fill value for missing values.
For example:
例如：
In [11]: df = pd.DataFrame(['hello', 'world'], columns=['words'])
In [12]: df
Out[12]:
words
0 hello
1 world
In [13]: df.words.str.contains(r'[hw]')
Out[13]:
0 True
1 True
Name: words, dtype: bool
In [14]: df.words.str.contains(r'he|wo')
Out[14]:
0 True
1 True
Name: words, dtype: bool
To count the occurences you can just sum this boolean Series:
要计算出现次数，您可以对这个布尔系列求和：
In [15]: df.words.str.contains(r'he|wo').sum()
Out[15]: 2
In [16]: df.words.str.contains(r'he').sum()
Out[16]: 1
回答by Dan Allan
To count the total number of matches, use s.str.match(...).str.get(0).count().
要计算匹配的总数，请使用s.str.match(...).str.get(0).count()。
If your regex will be matching several unique words, to be tallied individually, use
s.str.match(...).str.get(0).groupby(lambda x: x).count()
如果您的正则表达式将匹配几个独特的单词，要单独计算，请使用
s.str.match(...).str.get(0).groupby(lambda x: x).count()
It works like this:
它是这样工作的：
In [12]: s
Out[12]:
0 ax
1 ay
2 bx
3 by
4 bz
dtype: object
The matchstring method handles regular expressions...
该match字符串的方法处理正则表达式...
In [13]: s.str.match('(b[x-y]+)')
Out[13]:
0 []
1 []
2 (bx,)
3 (by,)
4 []
dtype: object
...but the results, as given, are not very convenient. The string method gettakes the matches as strings and converts empty results to NaNs...
...但结果，正如给定的，不是很方便。string 方法get将匹配项作为字符串并将空结果转换为 NaN...
In [14]: s.str.match('(b[x-y]+)').str.get(0)
Out[14]:
0 NaN
1 NaN
2 bx
3 by
4 NaN
dtype: object
...which are not counted.
……不计算在内。
In [15]: s.str.match('(b[x-y]+)').str.get(0).count()
Out[15]: 2
