value_counts omits NaN by default so you’re most likely dealing with “”.

So you can just filter them out like

filter = df["Tenant"] != ""
dfNew = df[filter]

You can use this variation:

import pandas as pd
vals = {
    'name' : ['n1', 'n2', 'n3', 'n4', 'n5', 'n6', 'n7'],
    'gender' : ['m', 'f', 'f', 'f',  'f', 'c', 'c'],
    'age' : [39, 12, 27, 13, 36, 29, 10],
    'education' : ['ma', None, 'school', None, 'ba', None, None]
}
df_vals = pd.DataFrame(vals) #converting dict to dataframe

This will output(** – highlighting only desired rows):

   age education gender name
0   39        ma      m   n1 **
1   12      None      f   n2    
2   27    school      f   n3 **
3   13      None      f   n4
4   36        ba      f   n5 **
5   29      None      c   n6
6   10      None      c   n7

So to drop everything that does not have an ‘education’ value, use the code below:

df_vals = df_vals[~df_vals['education'].isnull()] 

(‘~’ indicating NOT)

Result:

   age education gender name
0   39        ma      m   n1
2   27    school      f   n3
4   36        ba      f   n5

There’s a situation where the cell has white space, you can’t see it, use

df['col'].replace('  ', np.nan, inplace=True)

to replace white space as NaN, then

df= df.dropna(subset=['col'])

Pythonic + Pandorable: df[df['col'].astype(bool)]

Empty strings are falsy, which means you can filter on bool values like this:

df = pd.DataFrame({
    'A': range(5),
    'B': ['foo', '', 'bar', '', 'xyz']
})
df
   A    B
0  0  foo
1  1     
2  2  bar
3  3     
4  4  xyz

df['B'].astype(bool)                                                                                                                      
0     True
1    False
2     True
3    False
4     True
Name: B, dtype: bool

df[df['B'].astype(bool)]                                                                                                                  
   A    B
0  0  foo
2  2  bar
4  4  xyz

If your goal is to remove not only empty strings, but also strings only containing whitespace, use str.strip beforehand:

df[df['B'].str.strip().astype(bool)]
   A    B
0  0  foo
2  2  bar
4  4  xyz

Faster than you Think

.astype is a vectorised operation, this is faster than every option presented thus far. At least, from my tests. YMMV.

Here is a timing comparison, I’ve thrown in some other methods I could think of.

Benchmarking code, for reference:

import pandas as pd
import perfplot

df1 = pd.DataFrame({
    'A': range(5),
    'B': ['foo', '', 'bar', '', 'xyz']
})

perfplot.show(
    setup=lambda n: pd.concat([df1] * n, ignore_index=True),
    kernels=[
        lambda df: df[df['B'].astype(bool)],
        lambda df: df[df['B'] != ''],
        lambda df: df[df['B'].replace('', np.nan).notna()],  # optimized 1-col
        lambda df: df.replace({'B': {'': np.nan}}).dropna(subset=['B']),  
    ],
    labels=['astype', "!= ''", "replace + notna", "replace + dropna", ],
    n_range=[2**k for k in range(1, 15)],
    xlabel='N',
    logx=True,
    logy=True,
    equality_check=pd.DataFrame.equals)

If you don’t care about the columns where the missing files are, considering that the dataframe has the name New and one wants to assign the new dataframe to the same variable, simply run

New = New.drop_duplicates()

If you specifically want to remove the rows for the empty values in the column Tenant this will do the work

New = New[New.Tenant != '']

This may also be used for removing rows with a specific value – just change the string to the value that one wants.

Note: If instead of an empty string one has NaN, then

New = New.dropna(subset=['Tenant'])

Alternatively, you can use query.

• If your missing values are empty strings:

  df.query('Tenant != ""')
  

• If the missing values are NaN:

  df.query('Tenant == Tenant')
  

  (This works since np.nan != np.nan)

For anyone who reads data from a csv/tsv file which contains empty string cells, pandas will automatically convert them to NaN values (see the documentation). Assuming these cells are in column "c2", a way to filter them out is:

df[~df["c2"].isna()]

Note that the tilde operator does bitwise negation.