Python: How to drop a row whose particular column is empty/NaN?

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面向向阳花
面向向阳花 2020-12-01 11:00

I have a csv file. I read it:

import pandas as pd
data = pd.read_csv(\'my_data.csv\', sep=\',\')
data.head()

It has output like:

         


        
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  • 2020-12-01 11:25

    You can use the method dropna for this:

    data.dropna(axis=0, subset=('sms', ))
    

    See the documentation for more details on the parameters.

    Of course there are multiple ways to do this, and there are some slight performance differences. Unless performance is critical, I would prefer the use of dropna() as it is the most expressive.

    import pandas as pd
    import numpy as np
    
    i = 10000000
    
    # generate dataframe with a few columns
    df = pd.DataFrame(dict(
        a_number=np.random.randint(0,1e6,size=i),
        with_nans=np.random.choice([np.nan, 'good', 'bad', 'ok'], size=i),
        letter=np.random.choice(list('abcdefghijklmnop'), size=i))
                     )
    
    # using notebook %%timeit
    a = df.dropna(subset=['with_nans'])
    #1.29 s ± 112 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
    
    # using notebook %%timeit
    b = df[~df.with_nans.isnull()]
    #890 ms ± 59.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
    
    # using notebook %%timeit
    c = df.query('with_nans == with_nans')
    #1.71 s ± 100 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
    
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  • 2020-12-01 11:32

    Use dropna with parameter subset for specify column for check NaNs:

    data = data.dropna(subset=['sms'])
    print (data)
       id city department   sms  category
    1   2  lhr    revenue  good         1
    

    Another solution with boolean indexing and notnull:

    data = data[data['sms'].notnull()]
    print (data)
       id city department   sms  category
    1   2  lhr    revenue  good         1
    

    Alternative with query:

    print (data.query("sms == sms"))
       id city department   sms  category
    1   2  lhr    revenue  good         1
    

    Timings

    #[300000 rows x 5 columns]
    data = pd.concat([data]*100000).reset_index(drop=True)
    
    In [123]: %timeit (data.dropna(subset=['sms']))
    100 loops, best of 3: 19.5 ms per loop
    
    In [124]: %timeit (data[data['sms'].notnull()])
    100 loops, best of 3: 13.8 ms per loop
    
    In [125]: %timeit (data.query("sms == sms"))
    10 loops, best of 3: 23.6 ms per loop
    
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