Pandas条件计数实现：如何为车辆每日行程生成连续序号列

Hey there! Let's sort out this ride counting problem for you. First, let's break down why your current code isn't working, then I'll show you two reliable ways to get the exact result you want.

Why Your Current Code Fails #

• Wrong parameter in apply: When you use x[df["Is_first_ride"]], you're passing the entire Is_first_ride Series to the counter function instead of just the value from the current row. It should be x["Is_first_ride"] instead.
• State isn't preserved across rows: The counter function resets counter = 1 every time it's called (since apply runs it row-by-row). Even if you fixed the parameter, the while loop would only run once per row, so you'd never get a running count that carries over between rows.
• Ambiguous boolean check: Comparing a Series (from your incorrect parameter) to 1 creates a boolean Series, which pandas can't interpret as a single True/False value—hence the "Series is Ambiguous" error.

Solution 1: Use Your Existing Is_first_ride Column #

We can use the Is_first_ride flags to create grouping IDs, then count within each group:

# Create a unique group ID for each new day's first ride (each 1 in Is_first_ride starts a new group)
df['group_id'] = df['Is_first_ride'].cumsum()

# Generate the ride counter by counting positions within each group
df['#_of_ride'] = df.groupby('group_id').cumcount() + 1

# Optional: Drop the temporary group_id column if you don't need it
df = df.drop('group_id', axis=1)

Solution 2: Directly Group by Car and Date (Recommended) #

Since you're counting rides per car per day, grouping directly by car and date is more straightforward and reliable (it doesn't depend on the Is_first_ride column being perfectly accurate):

# Group by car and date, then count the position of each row in its group (add 1 to start from 1)
df['#_of_ride'] = df.groupby(['car', 'date']).cumcount() + 1

Test It With Your Sample Data #

Here's a full example using your sample input to confirm it works:

import pandas as pd

# Your sample data
data = {
    'car': ['Ford', 'Ford', 'Ford', 'Ford', 'Fiat', 'Fiat', 'Fiat', 'Fiat', 'Ford', 'Ford'],
    'date': ['20.1.2021']*4 + ['20.1.2021']*4 + ['21.1.2021']*2,
    'Is_first_ride': [1,0,0,0,1,0,0,0,1,0]
}
df = pd.DataFrame(data)

# Apply Solution 2 (the simpler one)
df['#_of_ride'] = df.groupby(['car', 'date']).cumcount() + 1

print(df)

This will output exactly the table you wanted:

car       date  Is_first_ride  #_of_ride
0   Ford  20.1.2021              1          1
1   Ford  20.1.2021              0          2
2   Ford  20.1.2021              0          3
3   Ford  20.1.2021              0          4
4   Fiat  20.1.2021              1          1
5   Fiat  20.1.2021              0          2
6   Fiat  20.1.2021              0          3
7   Fiat  20.1.2021              0          4
8   Ford  21.1.2021              1          1
9   Ford  21.1.2021              0          2

内容的提问来源于stack exchange，提问作者alex_loxn