Masking a Pandas DataFrame rows based on the whole row

pandas, python

Solution

The process of clarifying my question has lead me, in a roundabout way, to finding the answer. This question also helped point me in the right direction. Here's what I figured out:

import pandas as pd
# Set up my fake test data again. My actual data is described
# in the question.
cols = ['band1','band2','band3','band4','band5','band6','band7','band8']
rdf = pd.DataFrame(np.random.randint(0,10,80).reshape(10,8),columns=cols)
zdf = pd.DataFrame(np.zeros( (3,8) ),columns=cols)
df = pd.concat((zdf,rdf)).reset_index(drop=True)

# View the dataframe. (sorry about the alignment, I don't
# want to spend the time putting in all the spaces)
df

    band1   band2   band3   band4   band5   band6   band7   band8
0   0   0   0   0   0   0   0   0
1   0   0   0   0   0   0   0   0
2   0   0   0   0   0   0   0   0
3   6   3   7   0   1   7   1   8
4   9   2   6   8   7   1   4   3
5   4   2   1   1   3   2   1   9
6   5   3   8   7   3   7   5   2
7   8   2   6   0   7   2   0   7
8   1   3   5   0   7   3   3   5
9   1   8   6   0   1   5   7   7
10  4   2   6   2   2   2   4   9
11  8   7   8   0   9   3   3   0
12  6   1   6   8   2   0   2   5

13 rows × 8 columns

# This is essentially the same as item #2 under Fails
# in my question. It gives me the indexes of the rows
# I want unmasked as True and those I want masked as
# False. However, the result is not the right shape to
# use as a mask.
df.apply( lambda row: any([i<>0 for i in row]),axis=1 )
0     False
1     False
2     False
3      True
4      True
5      True
6      True
7      True
8      True
9      True
10     True
11     True
12     True
dtype: bool

# This is what actually works. By setting broadcast to
# True, I get a result that's the right shape to use.
land_rows = df.apply( lambda row: any([i<>0 for i in row]),axis=1, 
                      broadcast=True )

land_rows

Out[92]:
    band1   band2   band3   band4   band5   band6   band7   band8
0   0   0   0   0   0   0   0   0
1   0   0   0   0   0   0   0   0
2   0   0   0   0   0   0   0   0
3   1   1   1   1   1   1   1   1
4   1   1   1   1   1   1   1   1
5   1   1   1   1   1   1   1   1
6   1   1   1   1   1   1   1   1
7   1   1   1   1   1   1   1   1
8   1   1   1   1   1   1   1   1
9   1   1   1   1   1   1   1   1
10  1   1   1   1   1   1   1   1
11  1   1   1   1   1   1   1   1
12  1   1   1   1   1   1   1   1

13 rows × 8 columns

# This produces the result I was looking for:
df.where(land_rows)

Out[93]:
    band1   band2   band3   band4   band5   band6   band7   band8
0   NaN     NaN     NaN     NaN     NaN     NaN     NaN     NaN
1   NaN     NaN     NaN     NaN     NaN     NaN     NaN     NaN
2   NaN     NaN     NaN     NaN     NaN     NaN     NaN     NaN
3   6   3   7   0   1   7   1   8
4   9   2   6   8   7   1   4   3
5   4   2   1   1   3   2   1   9
6   5   3   8   7   3   7   5   2
7   8   2   6   0   7   2   0   7
8   1   3   5   0   7   3   3   5
9   1   8   6   0   1   5   7   7
10  4   2   6   2   2   2   4   9
11  8   7   8   0   9   3   3   0
12  6   1   6   8   2   0   2   5

13 rows × 8 columns

Thanks again to those who helped. Hopefully the solution I found will be of use to somebody at some point.

I found another way to do the same thing. There are more steps involved but, according to %timeit, it is about 9 times faster. Here it is:

def mask_all_zero_rows_numpy(df):
    """
    Take a dataframe, find all the rows that contain only zeros
    and mask them. Return a dataframe of the same shape with all
    Nan rows in place of the all zero rows.
    """
    no_data = -99
    arr = df.as_matrix().astype(int16)
    # make a row full of the 'no data' value
    replacement_row = np.array([no_data for x in range(arr.shape[1])], dtype=int16)
    # find out what rows are all zeros
    mask_rows = ~arr.any(axis=1)
    # replace those all zero rows with all 'no_data' rows
    arr[mask_rows] = replacement_row
    # create a masked array with the no_data value masked
    marr = np.ma.masked_where(arr==no_data,arr)
    # turn masked array into a data frame
    mdf = pd.DataFrame(marr,columns=df.columns)
    return mdf

The result of `mask_all_zero_rows_numpy(df)` should be the same as `Out[93]:` above.

Problem

Background: I'm working with 8 band multispectral satellite imagery and estimating water depth from reflectance values. Using statsmodels, I've come up with an OLS model that will predict depth for each pixel based on the 8 reflectance values of that pixel. In order to work easily with the OLS model, I've stuck all the pixel reflectance values into a pandas dataframe formated like the one in the example below; where each row represents a pixel and each column is a spectral band of the multispectral image. Due to some pre-processing steps, all the on-shore pixels have been transformed to all zeros. I don't want to try and predict the 'depth' of those pixels so I want to restrict my OLS model predictions to the rows that are NOT all zero values. I will need to reshape my results back to the row x col dimensions of the original image so I can't just drop the all zero rows. Specific Question: I've got a Pandas dataframe. Some rows contain all zeros. I would like to mask those rows for some calculations but I need to keep the rows. I can't figure out how to mask all the entries for rows that are all zero. For example: ``` In [1]: import pandas as pd In [2]: import numpy as np # my actual data has about 16 million rows so # I'll simulate some data for the example. In [3]: cols = ['band1','band2','band3','band4','band5','band6','band7','band8'] In [4]: rdf = pd.DataFrame(np.random.randint(0,10,80).reshape(10,8),columns=cols) In [5]: zdf = pd.DataFrame(np.zeros( (3,8) ),columns=cols) In [6]: df = pd.concat((rdf,zdf)).reset_index(drop=True) In [7]: df Out[7]: band1 band2 band3 band4 band5 band6 band7 band8 0 9 9 8 7 2 7 5 6 1 7 7 5 6 3 0 9 8 2 5 4 3 6 0 3 8 8 3 6 4 5 0 5 7 4 5 4 8 3 2 4 1 3 2 5 5 9 7 6 3 8 7 8 4 6 6 2 8 2 2 6 9 8 7 9 4 0 2 7 6 4 8 8 1 3 5 3 3 3 0 1 9 4 2 9 7 3 5 5 0 10 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 [13 rows x 8 columns] ``` I know I can get just the rows I'm interested in by doing this: ``` In [8]: df[df.any(axis=1)==True] Out[8]: band1 band2 band3 band4 band5 band6 band7 band8 0 9 9 8 7 2 7 5 6 1 7 7 5 6 3 0 9 8 2 5 4 3 6 0 3 8 8 3 6 4 5 0 5 7 4 5 4 8 3 2 4 1 3 2 5 5 9 7 6 3 8 7 8 4 6 6 2 8 2 2 6 9 8 7 9 4 0 2 7 6 4 8 8 1 3 5 3 3 3 0 1 9 4 2 9 7 3 5 5 0 [10 rows x 8 columns] ``` But I need to reshape the data again later so I'll need those rows to be in the right place. I've tried all sorts of things including `df.where(df.any(axis=1)==True)` but I can't find anything that works. Fails: `df.any(axis=1)==True` gives me `True` for the rows I'm interested in and `False` for rows I'd like to mask but when I try `df.where(df.any(axis=1)==True)` I just get back the whole data frame complete with all the zeros. I want the whole data frame but with all the values in those zero rows masked so, as I understand it, they should show up as Nan, right? I tried getting the indexes of the rows with all zeros and masking by row: ``` mskidxs = df[df.any(axis=1)==False].index df.mask(df.index.isin(mskidxs)) ``` That didn't work either that gave me: ``` ValueError: Array conditional must be same shape as self ``` The `.index` is just giving an `Int64Index` back. I need a boolean array the same dimensions as my data frame and I just can't figure out how to get one. Thanks in advance for your help. -Jared

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