Conditional check if-statement python-CodePudding

Let's say I have the following dataframe:

  df = pd.DataFrame({"a":[5, 6, 7, 8, 11],
                       "b":["A", "B", "C", "B", "A"],
                       "c":[27, 6, 1, 8, 3],
                       "d":[31, 26, 17, 8, 3],
                       "e":[12, np.nan, 11, 8, 6],
                       "f":[5, np.nan, 5, np.nan, 7],
                       "g":[27, 5, 12, 4, 19],
                       "h":[6, 16, 11, 2, 9],
                       "i":['One', "Two", "One", "Three", "One"]
                      })
    df

I want to check if there are rows who do not meet certain requirements. For example:

if ((df.d < df.c) & (df.a != 0) & (df.i == "X")).all() is False:
    raise ValueError(f"Incorrect! ")
else:
    print('Seems ok!')

This should raise a ValueError, as not in every row column d is larger than column c, and there is no value 'X' in column i. However, I keep getting the result 'Seems ok!'. What is wrong in the if-statement that is does not raise a ValueError?

CodePudding user response：

Use not or == False instead is:

if not ((df.d < df.c) & (df.a != 0) & (df.i == "X")).all() :
    raise ValueError(f"Incorrect! ")
else:
    print('Seems ok!')

Or:

if ((df.d < df.c) & (df.a != 0) & (df.i == "X")).all() :
    print('Seems ok!')
else:
    raise ValueError(f"Incorrect! ")

CodePudding user response：

Per De Morgan's law, you can invert all your conditions, use OR (|), and any:

if ((df.d >= df.c) | (df.a == 0) | (df.i != "X")).any():
    raise ValueError(f"Incorrect! ")
else:
    print('Seems ok!')

output: ValueError: Incorrect!