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)`

.

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
```