`TypeError` on conditional

[sivabudh]

Is there a way to make the following working Pandas code more concise?

value = data | px[px.SOME_COLUMN == 10] # data: DataFrame
if not value.empty:
    print(value.ANOTHER_COLUMN.astype(str).values[0])

I tried this:

data | px[px.SOME_COLUMN == 40] | (px if not px.empty else exit()) | px.ANOTHER_COLUMN.astype(str).values[0] | p(print)

but I got a TypeError:

Traceback (most recent call last):
  File "/test_pandas.py", line 25, in <module>
    data | px[px.SOME_COLUMN == 40] | (px if not px.empty else exit()) | px.ANOTHER_COLUMN.astype(str).values[0] | p(print)
TypeError: 'Pipe' object cannot be interpreted as an integer

Essentially, I'm trying not to print if the DataFrame is empty.

[mohammad7t]

Using px in ... if ...px... else ... is not supported. This error is because of the order python evaluates ifelse expressionss. The px.empty is evaluated before data | px[px.SOME_COLUMN == 40] is passed to it.

Alternatively, you can use p:

data | px[px.SOME_COLUMN == 40] | p(lambda x: x if not x.empty else exit()) | px.ANOTHER_COLUMN.astype(str).values[0] | p(print)

[GitHunter0]

data | px[px.SOME_COLUMN == 40] | p(lambda x: x if not x.empty else exit()) | px.ANOTHER_COLUMN.astype(str).values[0] | p(print)

@mhsekhavat , exit() shuts down the Jupyter kernel in VScode, I believe it would be better to use None in this concise syntax, what do you think?

import pandas as pd
df = pd.DataFrame({'col1': [1,2], 'col2': [3,4]})
print(df)

# Conventional syntax
dfc = df[df.col1==2]
if not dfc.empty:
    print(dfc.col2.astype("str").values[0])

# sspipe syntax
from sspipe import p, px
( df | px[px.col1==2] 
     | p( lambda x: print(x.col2.astype("str").values[0])
                    if not x.empty 
                    else None ) )

[mohammad7t]

@GitHunter0 you're right. Returning None makes more sense.