[英]Passing operators as functions to use with Pandas data frames
I am selecting data from series on basis of threshold . 我正在根据阈值从系列中选择数据。
>>> s = pd.Series(np.random.randn(5))
>>> s
0 -0.308855
1 -0.031073
2 0.872700
3 -0.547615
4 0.633501
dtype: float64
>>> cfg = {'threshold' : 0 , 'op' : 'less' }
>>> ops = {'less' : '<', 'more': '>' , 'equal': '==' , 'not equal' : '!='}
>>> ops[cfg['op']]
'<'
>>> s[s < cfg['threshold']]
0 -0.308855
1 -0.031073
3 -0.547615
dtype: float64
I want to use ops[cfg['op']] in last line of code , instead of '<'. 我想在代码的最后一行中使用ops [cfg ['op']],而不要使用'<'。 I am willing to change key , values of ops dict if required (like -lt instead of <). 如果愿意,我愿意更改key,ops dict的值(例如-lt代替<)。 How this can be done? 如何做到这一点?
I'm all about @cᴏʟᴅsᴘᴇᴇᴅ's answer and @Zero's linked Q&A... 我全都关心@cᴏʟᴅsᴘᴇᴇᴅ的答案和@Zero的相关问答。
But here is an alternative with numexpr
但是这是numexpr
的替代numexpr
import numexpr as ne
s[ne.evaluate('s {} {}'.format(ops[cfg['op']], cfg['threshold']))]
0 -0.308855
1 -0.031073
3 -0.547615
Name: A, dtype: float64
I reopened this question after having been closed as a dup of How to pass an operator to a python function? 在作为如何将运算符传递给python函数的dup封闭后,我重新打开了这个问题。
The question and answers are great and I showed my appreciation with up votes. 问题和答案都很棒,我对投票表示赞赏。
Asking in the context of a pandas.Series
opens it up to using answers that include numpy
and numexpr
. 在pandas.Series
上下文中pandas.Series
可以使用包括numpy
和numexpr
答案进行numexpr
。 Whereas trying to answer the dup target with this answer would be pure nonsense. 而试图用这个答案来回答重复目标将是纯粹的废话。
Define a dictionary of methods that can stand in for your operators. 定义可以代表您的操作员的方法的词典。
import operator
d = {
'more' : operator.gt,
'less' : operator.lt,
'equal' : operator.eq,
'not equal' : operator.ne
}
Now, just index into your dictionary and apply your function parameters. 现在,只需索引到字典中并应用函数参数即可。
m = d[cfg['op']](s, cfg['threshold'])
m
0 False
1 True
2 True
3 False
4 False
dtype: bool
s[m]
1 -0.262054
2 -1.300810
dtype: float64
Here, 这里,
d[cfg['op']](s, cfg['threshold'])
Is translated into 被翻译成
operator.lt(s, 0)
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