python – Remove rows not .isin('X')-ThrowExceptions

Exception or error:

Sorry just getting into Pandas, this seems like it should be a very straight forward question. How can I use the isin('X') to remove rows that are in the list X? In R I would write !which(a %in% b).

How to solve:

You have many options. Collating some of the answers above and the accepted answer from this post you can do:
1. df[-df["column"].isin(["value"])]
2. df[~df["column"].isin(["value"])]
3. df[df["column"].isin(["value"]) == False]
4. df[np.logical_not(df["column"].isin(["value"]))]

Note: for option 4 for you’ll need to import numpy as np

Answer:

You can use numpy.logical_not to invert the boolean array returned by isin:

In [63]: s = pd.Series(np.arange(10.0))

In [64]: x = range(4, 8)

In [65]: mask = np.logical_not(s.isin(x))

In [66]: s[mask]
Out[66]: 
0    0
1    1
2    2
3    3
8    8
9    9

As given in the comment by Wes McKinney you can also use

s[~s.isin(x)]

Answer:

All you have to do is create a subset of your dataframe where the isin method evaluates to False:

df = df[df['Column Name'].isin(['Value']) == False]

Answer:

You can use the DataFrame.select method:

In [1]: df = pd.DataFrame([[1,2],[3,4]], index=['A','B'])

In [2]: df
Out[2]: 
   0  1
A  1  2
B  3  4

In [3]: L = ['A']

In [4]: df.select(lambda x: x in L)
Out[4]: 
   0  1
A  1  2

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