pd.Index.get_indexer

We can use pd.Index.get_indexer to get integer index.

idx = df.index.get_indexer(list_of_target_labels)
# If you only have single label we can use tuple unpacking here.
[idx] = df.index.get_indexer([country_name])

NB: pd.Index.get_indexer takes a list and returns a list. Integers from 0 to n - 1 indicating that the index at these positions matches the corresponding target values. Missing values in the target are marked by -1.

np.where

You could also use np.where here.

idx = np.where(df.index == country_name)[0]

list.index

We could also use list.index after converting Pd.Index to list using pd.Index.tolist

idx = df.index.tolist().index(country_name)

We can use Index.get_indexer:

df.index.get_indexer(['Peru'])

[3]

Or we can build a RangeIndex based on the size of the DataFrame then subset that instead:

pd.RangeIndex(len(df))[df.index == 'Peru']

Int64Index([3], dtype='int64')

Since we're only looking for a single label and the indexes are "all unique" we can also use Index.get_loc:

df.index.get_loc('Peru')

3

Sample DataFrame:

import pandas as pd

df = pd.DataFrame({
    'A': [1, 2, 3, 4, 5]
}, index=['Bahamas', 'Cameroon', 'Ecuador', 'Peru', 'Japan'])

df:

          A
Bahamas   1
Cameroon  2
Ecuador   3
Peru      4
Japan     5

Why you don make the index to be created with numbers instead of text? Because your df can be sorted in many ways beyond the alphabetical, and you can lose the rows count. With numbered index this wouldn't be a problem.