Python pandas: how to remove nan and -inf values

Question

I have the following dataframe

           time       X    Y  X_t0     X_tp0  X_t1     X_tp1  X_t2     X_tp2
0         0.002876    0   10     0       NaN   Na         


        

Answer 1

df.replace only replaces the first occurrence on the value and thus the error

df = list(filter(lambda x: x!= inf, df)) would remove all occurrences of inf and then the drop function can be used

Answer 2

I prefer to set the options so that inf values are calculated to nan;

s1 = pd.Series([0, 1, 2])
s2 = pd.Series([2, 1, 0])
s1/s2
# Outputs:
# 0.0
# 1.0
# inf
# dtype: float64

pd.set_option('mode.use_inf_as_na', True)
s1/s2
# Outputs:
# 0.0
# 1.0
# NaN
# dtype: float64

Note you can also use context;

with pd.option_context('mode.use_inf_as_na', True):
    print(s1/s2)
# Outputs:
# 0.0
# 1.0
# NaN
# dtype: float64

Answer 3

You can replace inf and -inf with NaN, and then select non-null rows.

df[df.replace([np.inf, -np.inf], np.nan).notnull().all(axis=1)]  # .astype(np.float64) ?

or

df.replace([np.inf, -np.inf], np.nan).dropna(axis=1)

Check the type of your columns returns to make sure they are all as expected (e.g. np.float32/64) via df.info().

Answer 4

Use pd.DataFrame.isin and check for rows that have any with pd.DataFrame.any. Finally, use the boolean array to slice the dataframe.

df[~df.isin([np.nan, np.inf, -np.inf]).any(1)]

             time    X    Y  X_t0     X_tp0   X_t1     X_tp1   X_t2     X_tp2
4        0.037389    3   10     3  0.333333    2.0  0.500000    1.0  1.000000
5        0.037393    4   10     4  0.250000    3.0  0.333333    2.0  0.500000
1030308  9.962213  256  268   256  0.000000  256.0  0.003906  255.0  0.003922

Answer 5

df.replace([np.inf, -np.inf], np.nan)

df.dropna(inplace=True)

Answer 6

Instead of dropping rows which contain any nulls and infinite numbers, it is more succinct to the reverse the logic of that and instead return the rows where all cells are finite numbers. The numpy isfinite function does this and the '.all(1)' will only return a TRUE if all cells in row are finite.

df = df[np.isfinite(df).all(1)]