如何使用Pandas按连续天数分组拆分DataFrame？

Great question! Let's break down how to group your DataFrame by consecutive blocks of days (whether it's 2 days, 4 days, or any custom interval) in Pandas. Here are two straightforward approaches depending on your needs:

1. Fixed Consecutive Day Groups (e.g., Every 2 Days) #

If you want to split days into equal-sized consecutive blocks (like every 2 days, every 4 days), the simplest way is to generate a group ID using integer division. This works regardless of how many rows each day has.

Step-by-Step Code: #

import pandas as pd

# Your original data setup
df = pd.DataFrame({
 'day' : [1,1,2,3,3,3,4,4],
 'user' : ['A','A','B','C','B','B','C','C'],
 'score': [10,5,5,10,5,5,0,5]
})
df["total"] = df.cumsum()["score"]

# Define your group size (e.g., 2 days per group, change to 4 for larger blocks)
days_per_group = 2

# Calculate group ID for each row
# (day - 1) ensures we start grouping from day 1 as the first block
df['group_id'] = (df['day'] - 1) // days_per_group

# Now you can group by this ID to get your desired blocks
for group_num, group_data in df.groupby('group_id'):
    start_day = group_num * days_per_group + 1
    end_day = (group_num + 1) * days_per_group
    print(f"=== Group {group_num + 1} (Days {start_day} to {end_day}) ===")
    print(group_data)
    print("\n")

How It Works: #

The expression (df['day'] - 1) // days_per_group maps consecutive days to the same group ID:

• For days_per_group=2: Day 1 & 2 → ID 0; Day 3 & 4 → ID 1
• For days_per_group=4: Days 1-4 → ID 0, and so on

If your days don't start at 1 (e.g., starting at day 5), adjust the formula to (df['day'] - df['day'].min()) // days_per_group to maintain consecutive blocks relative to your first day.

2. Custom Day Interval Groups (Arbitrary Ranges) #

If you need non-uniform groups (like 1-3 days in one group, 4-6 in another), use pd.cut() to define custom day boundaries.

Step-by-Step Code: #

# Define custom day bins (adjust these to your needs)
# Bins are [lower bound, upper bound), right=False makes intervals left-closed
day_bins = [0, 2, 4, float('inf')]
group_labels = ['Group 1 (Days 1-2)', 'Group 2 (Days 3-4)', 'Group 3+']

# Assign groups to each row
df['custom_group'] = pd.cut(df['day'], bins=day_bins, labels=group_labels, right=False)

# View grouped results
print(df.groupby('custom_group').apply(lambda x: x))

How It Works: #

• pd.cut() splits the day column into the intervals you define. Setting right=False ensures intervals are left-closed (e.g., 0 < day ≤ 2 captures days 1 and 2).
• You can tweak day_bins and group_labels to match any custom grouping logic you need.

内容的提问来源于stack exchange，提问作者Shlomi Schwartz