Using np.where but maintaining exisitng values if condition is False

There is a pandas.Series method (where incidentally) for exactly this kind of task. It seems a little backward at first, but from the documentation.

Series.where(cond, other=nan, inplace=False, axis=None, level=None, try_cast=False, raise_on_error=True)

Return an object of same shape as self and whose corresponding entries are from self where cond is True and otherwise are from other.

So, your example would become

cols = ['a', 'b', 'c', 'd']condition = (DF[cols] == 0).all(axis=1)for col in cols:    DF[col].where(~condition, np.nan, inplace=True)

But, if all you're trying to do is replace rows of all zeros for specific set of columns with NA, you could do this instead

DF.loc[condition, cols] = NA

EDIT

To answer your original question, np.where follows the same broadcasting rules as other array operations so you would replace ??? with DF[col], changing your example to:

cols = ['a', 'b', 'c', 'd']condition = (DF[cols] == 0).all(axis=1)for col in cols:    DF[col] = np.where(condition, NA, DF[col])

Proposed solutions work but for numpy array there is a simpler way without using DataFrame.

A solution would be :np_array[np.where(condition)] = value_of_condition_true_rows

You can do something like this:

    array_binary = np.where(array[i]<threshold,0,1)    array_sparse = np.multiply(array_binary,np.ones_like(array))

do an element-wise multiplication of the binary array and an array of ones using np.multiply. Hence, the non-zero elements will be recovered/maintained.array_sparse is the sparse version of array