pandas - find first occurrence

idxmax and argmax will return the position of the maximal value or the first position if the maximal value occurs more than once.

use idxmax on df.A.ne('a')

df.A.ne('a').idxmax()

3

or the numpy equivalent

(df.A.values != 'a').argmax()

3

However, if A has already been sorted, then we can use searchsorted

df.A.searchsorted('a', side='right')

array([3])

Or the numpy equivalent

df.A.values.searchsorted('a', side='right')

3

I found there is first_valid_index function for Pandas DataFrames that will do the job, one could use it as follows:

df[df.A!='a'].first_valid_index()

3

However, this function seems to be very slow. Even taking the first index of the filtered dataframe is faster:

df.loc[df.A!='a','A'].index[0]

Below I compare the total time(sec) of repeating calculations 100 times for these two options and all the codes above:

                      total_time_sec    ratio wrt fastest algo
searchsorted numpy:        0.0007        1.00
argmax numpy:              0.0009        1.29
for loop:                  0.0045        6.43
searchsorted pandas:       0.0075       10.71
idxmax pandas:             0.0267       38.14
index[0]:                  0.0295       42.14
first_valid_index pandas:  0.1181      168.71

Notice numpy's searchsorted is the winner and first_valid_index shows worst performance. Generally, numpy algorithms are faster, and the for loop does not do so bad, but it's just because the dataframe has very few entries.

For a dataframe with 10,000 entries where the desired entries are closer to the end the results are different, with searchsorted delivering the best performance:

                     total_time_sec ratio wrt fastest algo
searchsorted numpy:        0.0007       1.00
searchsorted pandas:       0.0076      10.86
argmax numpy:              0.0117      16.71
index[0]:                  0.0815     116.43
idxmax pandas:             0.0904     129.14
first_valid_index pandas:  0.1691     241.57
for loop:                  9.6504   13786.29

The code to produce these results is below:

import timeit

# code snippet to be executed only once 
mysetup = '''import pandas as pd
import numpy as np
df = pd.DataFrame({"A":['a','a','a','b','b'],"B":[1]*5})
'''

# code snippets whose execution time is to be measured   
mycode_set = ['''
df[df.A!='a'].first_valid_index()
''']
message = ["first_valid_index pandas:"]

mycode_set.append( '''df.loc[df.A!='a','A'].index[0]''')
message.append("index[0]: ")

mycode_set.append( '''df.A.ne('a').idxmax()''')
message.append("idxmax pandas: ")

mycode_set.append(  '''(df.A.values != 'a').argmax()''')
message.append("argmax numpy: ")

mycode_set.append( '''df.A.searchsorted('a', side='right')''')
message.append("searchsorted pandas: ")

mycode_set.append( '''df.A.values.searchsorted('a', side='right')''' )
message.append("searchsorted numpy: ")

mycode_set.append( '''for index in range(len(df['A'])):
    if df['A'][index] != 'a':
        ans = index
        break
        ''')
message.append("for loop: ")

total_time_in_sec = []
for i in range(len(mycode_set)):
    mycode = mycode_set[i]
    total_time_in_sec.append(np.round(timeit.timeit(setup = mysetup,\
         stmt = mycode, number = 100),4))

output = pd.DataFrame(total_time_in_sec, index = message, \
                      columns = ['total_time_sec' ])
output["ratio wrt fastest algo"] = \
np.round(output.total_time_sec/output["total_time_sec"].min(),2)

output = output.sort_values(by = "total_time_sec")
display(output)

For the larger dataframe:

mysetup = '''import pandas as pd
import numpy as np
n = 10000
lt = ['a' for _ in range(n)]
b = ['b' for _ in range(5)]
lt[-5:] = b
df = pd.DataFrame({"A":lt,"B":[1]*n})
'''

Using pandas groupby() to group by column or list of columns. Then first() to get the first value in each group.

import pandas as pd

df = pd.DataFrame({"A":['a','a','a','b','b'],
                   "B":[1]*5})

#Group df by column and get the first value in each group                   
grouped_df = df.groupby("A").first()

#Reset indices to match format
first_values = grouped_df.reset_index()

print(first_values)
>>>    A  B
   0   a  1
   1   b  1

For multiple conditions:

Let's say we have:

s = pd.Series(['a', 'a', 'c', 'c', 'b', 'd'])

And we want to find the first item different than a and c, we do:

n = np.logical_and(s.values != 'a', s.values != 'c').argmax()

Times:

import numpy as np
import pandas as pd
from datetime import datetime

ITERS = 1000

def pandas_multi_condition(s):
    ts = datetime.now()
    for i in range(ITERS):
        n = s[(s != 'a') & (s != 'c')].index[0]
    print(n)
    print(datetime.now() - ts)

def numpy_bitwise_and(s):
    ts = datetime.now()
    for i in range(ITERS):
        n = np.logical_and(s.values != 'a', s.values != 'c').argmax()
    print(n)
    print(datetime.now() - ts)

s = pd.Series(['a', 'a', 'c', 'c', 'b', 'd'])

print('pandas_multi_condition():')
pandas_multi_condition(s)
print()
print('numpy_bitwise_and():')
numpy_bitwise_and(s)

Output:

pandas_multi_condition():
4
0:00:01.144767

numpy_bitwise_and():
4
0:00:00.019013