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To: python-list@python.org
From: Peter Otten <__peter__@web.de>
Subject: Re: eval [was Re: dict to boolean expression, how to?]
Date: Sat, 02 Aug 2014 09:43:54 +0200
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Xref: csiph.com comp.lang.python:75522

Steven D'Aprano wrote:

> On Fri, 01 Aug 2014 17:44:27 +0200, Peter Otten wrote:
> [...]
>>> bool = ((df['a'] == 1) & (df['A'] == 0) |
>>>          (df['b'] == 1) & (df['B'] == 0) |
>>>          (df['c'] == 1) & (df['C'] == 0))
>>  
>> This is how it might look without eval():
>> 
>> #untested
>> result = functools.reduce(operator.or_, ((v == 1) & (df[k.upper()] == 0)
>> for k, v in df.items() if k.islower()))
> 
> For those who agree with Guido that reduce makes code unreadable:
> 
> result = True
> for key in df:
>     if key.islower():
>         result = result or (df[key] == 1 and df[key.upper()] == 0)

I cheated a bit and gave the solution that the OP was unlikely to come up 
with ;)


> Or if you insist on a single expression:
> 
> result = any(df[k] == 1 and df[k.upper()] == 0 for k in df if k.islower())
> 
> 
>> And here is an eval-based solution:
>> 
>> # untested
>> expr = "|".join(
>>     "((df[{}] == 1) | (df[{}] == 0))".format(c, c.upper()) for c in df
>>     is c.islower())
>> result = eval(expr)
> 
> I really don't believe that there is any benefit to that in readability,
> power, flexibility, or performance. 

You can put in a print(expr) to verify that it's identical to the original 
spelt out expression.

> Also, you're using bitwise operators
> instead of shortcut bool operators. Any reason why?

Since he started the thread Alex has moved the goal posts a bit and added 
that he is using a pandas DataFrame. That works much like a numpy array. So:

>>> df
   a  A    b     B   c   C
0  1  4  100  1000   7  14
1  2  1  200  2000  21  28
2  1  0  300  3000  35  42
>>> df["a"] == 1
0     True
1    False
2     True
Name: a, dtype: bool
>>> (df["a"] == 1) and (df["A"] == 0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: The truth value of an array with more than one element is 
ambiguous. Use a.any() or a.all()
>>> (df["a"] == 1) & (df["A"] == 0)
0    False
1    False
2     True
dtype: bool