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
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