Python: join string list if else oneliner

Question

Working with Pandas, I have to rewrite queries implemented as a dict:

query = {"height": 175}

The key is the attribute for the query and the value could be a scalar or iterable. In the first part I check if the value is not NaN and scalar. If this condition holds I write the query expression with the == symbol, but else if the value is Iterable I would need to write the expression with the in keyword. This is the actual code that I need to fix in order to work also with Iterables.

import numpy as np
from collections import Iterable


def query_dict_to_expr(query: dict) -> str:
    expr = " and ".join(["{} == {}"
                        .format(k, v) for k, v in query.items()
                         if (not np.isnan(v)
                             and np.isscalar(v))
                         else "{} in @v".format(k) if isinstance(v, Iterable)
                         ]
                        )
    return expr

but I got invalid syntax in correspondence with the else statement.

Possible duplicate of if/else in Python's list comprehension? — Right leg
– Right leg, Commented Aug 29, 2017 at 8:19
else "{} in @v".format(k) if isinstance(v, Iterable). Why the second if there? — 10101010
– 10101010, Commented Aug 29, 2017 at 8:20
If you can't work out how to write a comprehension, then write it out as a loop, that will do no harm unless performance is absolutely critical and will be more readable for you no doubt — Chris_Rands
– Chris_Rands, Commented Aug 29, 2017 at 8:22
to check if the value is Iterable. It has to be an else if statement. — Juan Carlos
– Juan Carlos, Commented Aug 29, 2017 at 8:23
You can't put if again after the else. The list comprehension can't get it is a nested if/else clause. — cezar
– cezar, Commented Aug 29, 2017 at 8:24

MaxU - stand with Ukraine · Accepted Answer · 2017-08-29 08:31:21Z

1

If I understand correctly, you don't need to check the type:

In [47]: query
Out[47]: {'height': 175, 'lst_col': [1, 2, 3]}

In [48]: ' and '.join(['{} == {}'.format(k,v) for k,v in query.items()])
Out[48]: 'height == 175 and lst_col == [1, 2, 3]'

Demo:

In [53]: df = pd.DataFrame(np.random.randint(5, size=(5,3)), columns=list('abc'))

In [54]: df
Out[54]:
   a  b  c
0  0  0  3
1  4  2  4
2  2  2  3
3  0  1  0
4  0  4  1

In [55]: query = {"a": 0, 'b':[0,4]}

In [56]: q = ' and '.join(['{} == {}'.format(k,v) for k,v in query.items()])

In [57]: q
Out[57]: 'a == 0 and b == [0, 4]'

In [58]: df.query(q)
Out[58]:
   a  b  c
0  0  0  3
4  0  4  1

edited Aug 29, 2017 at 8:31

answered Aug 29, 2017 at 8:26

MaxU - stand with Ukraine

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Comments

tobias_k · Accepted Answer · 2017-08-29 08:34:54Z

1

You misplaces the if/else in the comprehension. If you put the if after the for, like f(x) for x in iterable if g(x), this will filter the elements of the iterable (and can not be combined with an else). Instead, you want to keep all the elements, i.e. use f(x) for x in iterable where f(x) just happens to be a ternary expression, i.e. in the form a(x) if c(x) else b(x).

Instead, try like this (simplified non-numpy example):

>>> query = {"foo": 42, "bar": [1,2,3]}  
>>> " and ".join(["{} == {}".format(k, v)
                  if not isinstance(v, list)
                  else "{} in {}".format(k, v)
                  for k, v in query.items()])
'foo == 42 and bar in [1, 2, 3]'

edited Aug 29, 2017 at 8:34

answered Aug 29, 2017 at 8:29

tobias_k

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