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

Question

Assume an easy dataframe, for example

    A         B
0   1  0.810743
1   2  0.595866
2   3  0.154888
3   4  0.472721
4   5  0.894525
5   6  0.978174
6   7  0.859449
7   8  0.541247
8   9  0.232302
9  10  0.276566

How can I retrieve an index value of a row, given a condition?
For example:
dfb = df[df['A']==5].index.values.astype(int)
returns [4], but what I would like to get is just 4. This is causing me troubles later in the code.

Based on some conditions, I want to have a record of the indexes where that condition is fulfilled, and then select rows between.

I tried

dfb = df[df['A']==5].index.values.astype(int)
dfbb = df[df['A']==8].index.values.astype(int)
df.loc[dfb:dfbb,'B']

for a desired output

    A         B
4   5  0.894525
5   6  0.978174
6   7  0.859449

but I get TypeError: '[4]' is an invalid key

Asked By: durbachit

||

Source

Answer 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

Answered By: jezrael

Answer 2

The nature of wanting to include the row where A == 5 and all rows upto but not including the row where A == 8 means we will end up using iloc (loc includes both ends of slice).

In order to get the index labels we use idxmax. This will return the first position of the maximum value. I run this on a boolean series where A == 5 (then when A == 8) which returns the index value of when A == 5 first happens (same thing for A == 8).

Then I use searchsorted to find the ordinal position of where the index label (that I found above) occurs. This is what I use in iloc.

i5, i8 = df.index.searchsorted([df.A.eq(5).idxmax(), df.A.eq(8).idxmax()])
df.iloc[i5:i8]

numpy

you can further enhance this by using the underlying numpy objects the analogous numpy functions. I wrapped it up into a handy function.

def find_between(df, col, v1, v2):
    vals = df[col].values
    mx1, mx2 = (vals == v1).argmax(), (vals == v2).argmax()
    idx = df.index.values
    i1, i2 = idx.searchsorted([mx1, mx2])
    return df.iloc[i1:i2]

find_between(df, 'A', 5, 8)

timing

Answered By: piRSquared

Answer 3

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()

Answered By: dmdip

Answer 4

Little sum up for searching by row:

This can be useful if you don’t know the column values or if columns have non-numeric values

if u want get index number as integer u can also do:

item = df[4:5].index.item()
print(item)
4

it also works in numpy / list:

numpy = df[4:7].index.to_numpy()[0]
lista = df[4:7].index.to_list()[0]

in [x] u pick number in range [4:7], for example if u want 6:

numpy = df[4:7].index.to_numpy()[2]
print(numpy)
6

for DataFrame:

df[4:7]

    A          B
4   5   0.894525
5   6   0.978174
6   7   0.859449

or:

df[(df.index>=4) & (df.index<7)]

    A          B
4   5   0.894525
5   6   0.978174
6   7   0.859449

Answered By: JJx

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

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