Remove non-numeric rows in one column with pandas

Using pd.to_numeric

In [1079]: df[pd.to_numeric(df['id'], errors='coerce').notnull()]
Out[1079]:
  id  name
0  1     A
1  2     B
2  3     C
4  4     E
5  5     F

You could use standard method of strings isnumeric and apply it to each value in your id column:

import pandas as pd
from io import StringIO

data = """
id,name
1,A
2,B
3,C
tt,D
4,E
5,F
de,G
"""

df = pd.read_csv(StringIO(data))

In [55]: df
Out[55]: 
   id name
0   1    A
1   2    B
2   3    C
3  tt    D
4   4    E
5   5    F
6  de    G

In [56]: df[df.id.apply(lambda x: x.isnumeric())]
Out[56]: 
  id name
0  1    A
1  2    B
2  3    C
4  4    E
5  5    F

Or if you want to use id as index you could do:

In [61]: df[df.id.apply(lambda x: x.isnumeric())].set_index('id')
Out[61]: 
   name
id     
1     A
2     B
3     C
4     E
5     F

Edit. Add timings

Although case with pd.to_numeric is not using apply method it is almost two times slower than with applying np.isnumeric for str columns. Also I add option with using pandas str.isnumeric which is less typing and still faster then using pd.to_numeric. But pd.to_numeric is more general because it could work with any data types (not only strings).

df_big = pd.concat([df]*10000)

In [3]: df_big = pd.concat([df]*10000)

In [4]: df_big.shape
Out[4]: (70000, 2)

In [5]: %timeit df_big[df_big.id.apply(lambda x: x.isnumeric())]
15.3 ms ± 2.02 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [6]: %timeit df_big[df_big.id.str.isnumeric()]
20.3 ms ± 171 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [7]: %timeit df_big[pd.to_numeric(df_big['id'], errors='coerce').notnull()]
29.9 ms ± 682 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Given that df is your dataframe,

import numpy as np
df[df['id'].apply(lambda x: isinstance(x, (int, np.int64)))]

What it does is passing each value in the id column to the isinstance function and checks if it's an int. Then it returns a boolean array, and finally returning only the rows where there is True.

If you also need to account for float values, another option is:

import numpy as np
df[df['id'].apply(lambda x: type(x) in [int, np.int64, float, np.float64])]

Note that either way is not inplace, so you will need to reassign it to your original df, or create a new one:

df = df[df['id'].apply(lambda x: type(x) in [int, np.int64, float, np.float64])]
# or
new_df = df[df['id'].apply(lambda x: type(x) in [int, np.int64, float, np.float64])]