Get index of a row of a pandas dataframe as an integer

Solution 1:

The easier is add [0] - select first value of list with one element:

dfb = df[df['A']==5].index.values.astype(int)[0]
dfbb = df[df['A']==8].index.values.astype(int)[0]

dfb = int(df[df['A']==5].index[0])
dfbb = int(df[df['A']==8].index[0])

But if possible some values not match, error is raised, because first value not exist.

Solution is use next with iter for get default parameetr if values not matched:

dfb = next(iter(df[df['A']==5].index), 'no match')
print (dfb)
4

dfb = next(iter(df[df['A']==50].index), 'no match')
print (dfb)
no match

Then it seems need substract 1:

print (df.loc[dfb:dfbb-1,'B'])
4    0.894525
5    0.978174
6    0.859449
Name: B, dtype: float64

Another solution with boolean indexing or query:

print (df[(df['A'] >= 5) & (df['A'] < 8)])
   A         B
4  5  0.894525
5  6  0.978174
6  7  0.859449

print (df.loc[(df['A'] >= 5) & (df['A'] < 8), 'B'])
4    0.894525
5    0.978174
6    0.859449
Name: B, dtype: float64

print (df.query('A >= 5 and A < 8'))
   A         B
4  5  0.894525
5  6  0.978174
6  7  0.859449

Solution 2:

To answer the original question on how to get the index as an integer for the desired selection, the following will work :

df[df['A']==5].index.item()