I have a pandas DataFrame, with many NAN values in it.
How can I drop columns such that number_of_na_values > 2000?
I tried to do it like that:
toRemove = set()
naNumbersPerColumn = df.isnull().sum()
for i in naNumbersPerColumn.index:
if(naNumbersPerColumn[i]>2000):
toRemove.add(i)
for i in toRemove:
df.drop(i, axis=1, inplace=True)
Is there a more elegant way to do it?
Here's another alternative to keep the columns that have less than or equal to the specified number of nans in each column:
max_number_of_nas = 3000
df = df.loc[:, (df.isnull().sum(axis=0) <= max_number_of_nas)]
In my tests this seems to be slightly faster than the drop columns method suggested by Jianxun Li in the cases I tested (as shown below). However, I should note that the performance becomes more similar if you simply don't use the apply method (e.g. df.drop(df.columns[df.isnull().sum(axis=0) > max_number_of_nans], axis=1)). Just a reminder that when it comes to performance in pandas vectorization almost always wins out over apply.
np.random.seed(0)
df = pd.DataFrame(np.random.randn(10000,5), columns=list('ABCDE'))
df[df < 0] = np.nan
max_number_of_nans = 5010
%timeit c = df.loc[:, (df.isnull().sum(axis=0) <= max_number_of_nans)]
>> 1.1 ms ± 4.08 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit c = df.drop(df.columns[df.isnull().sum(axis=0) > max_number_of_nans], axis=1)
>> 1.3 ms ± 11.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit c = df.drop(df.columns[df.apply(lambda col: col.isnull().sum() > max_number_of_nans)], axis=1)
>> 2.11 ms ± 29.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Performance often varies with data size so don't forget to check whatever case is closest to your data.
np.random.seed(0)
df = pd.DataFrame(np.random.randn(10, 5), columns=list('ABCDE'))
df[df < 0] = np.nan
max_number_of_nans = 5
%timeit c = df.loc[:, (df.isnull().sum(axis=0) <= max_number_of_nans)]
>> 755 µs ± 4.84 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit c = df.drop(df.columns[df.isnull().sum(axis=0) > max_number_of_nans], axis=1)
>> 777 µs ± 12 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit c = df.drop(df.columns[df.apply(lambda col: col.isnull().sum() > max_number_of_nans)], axis=1)
>> 1.71 ms ± 17.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)