how can i make a string comparision using numexpr evaluate? not df.eval

Question

I'm trying to use Numexpr to make a filter.when i trying to run the next code snippet:

>>> import pandas as pd
>>> import numexpr
>>> df=pd.DataFrame({'a':[1,2,3],'b':['one','two','three']})
>>> df
   a      b
0  1    one
1  2    two
2  3  three
>>> a=df.a.values
>>> b=df.b.values
>>> numexpr.evaluate('a<2')
array([ True, False, False])
>>> numexpr.evaluate('b=="one"')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python36-64\lib\site-packages\numexpr\necompiler.py", line 822, in evaluate
    zip(names, arguments)]
  File "C:\Python36-64\lib\site-packages\numexpr\necompiler.py", line 821, in <listcomp>
    signature = [(name, getType(arg)) for (name, arg) in
  File "C:\Python36-64\lib\site-packages\numexpr\necompiler.py", line 703, in getType
    raise ValueError("unknown type %s" % a.dtype.name)
ValueError: unknown type object

for some reason,i can't use df.eval('b=="one"').Maybe I'm doing it the wrong way. Any help would be appreciated.

because i first use some code like df.query('a==1 and b="one"') ,it's slowly.so i want use numexpr speed up it — Grayhawk Rambo, Jul 23 '19 at 09:26
Just to be sure: do you really _need_ the string representation of your comparison terms or do you simply want to do that comparison? I.e. what about simply calculating `a < 2` and `b == 'one'`? — SpghttCd, Jul 23 '19 at 09:27
Are you aware of syntax like `df[(df.a<=2) & (df.b=='one')]`? — SpghttCd, Jul 23 '19 at 09:35
yes,i need the string comparision Because the filter condition is dynamic splic — Grayhawk Rambo, Jul 23 '19 at 09:47
Sorry, I don't know what _dynamic splic_ is. However, did you have a look at `eval` (if safety is not relevant in your application) `ast` or `asteval`? — SpghttCd, Jul 23 '19 at 10:39

how can i make a string comparision using numexpr evaluate? not df.eval

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