Python实现重复行计数：其他行重复≥3次时重置计数器

问题描述 #

现有如下DataFrame：

import pandas as pd
df = pd.DataFrame({"word": ["A","A","A","A","A","A","A","A","A","B","A","B","B","A","A","A","B","B","B","A","B","B"]})

需要实现的规则：

• 对每个word（A和B）维护一个累计计数器
• 每当另一个word连续出现3次及以上时，当前word的计数器重置为1重新累计
• 若另一个word仅连续出现1-2次，当前word的计数器继续累计，不重置

你尝试的普通分组累计代码（如下）无法满足上述规则，因为它只会单纯按word分组累计，不会根据另一个word的连续出现次数重置计数器：

df['num_aux']=1
df['cumulative_numero']=df.groupby(['word'])['num_aux'].cumsum()

解决方案 #

通过识别连续相同word的分组、判断重置触发条件，再结合复合分组实现需求：

import pandas as pd
import numpy as np

# 初始化DataFrame
df = pd.DataFrame({"word": ["A","A","A","A","A","A","A","A","A","B","A","B","B","A","A","A","B","B","B","A","B","B"]})

# 1. 标记连续相同word的分组
df['group'] = (df['word'] != df['word'].shift()).cumsum()

# 2. 计算每个连续分组的长度
group_lengths = df.groupby('group')['word'].count()
df['group_len'] = df['group'].map(group_lengths)

# 3. 标记重置触发点：当前行属于新word组，且前一个组（另一个word）的长度>=3
df['reset_trigger'] = (df['word'] != df['word'].shift()) & (df['group_len'].shift() >= 3)

# 4. 按word + 重置触发累计值分组，实现计数器的重置累计
df['num_aux'] = 1
df['cumulative_numero'] = df.groupby(['word', df['reset_trigger'].cumsum()])['num_aux'].cumsum()

# 可选：删除中间辅助列
df = df.drop(['group', 'group_len', 'reset_trigger', 'num_aux'], axis=1)

print(df)

结果说明 #

最终输出的cumulative_numero列符合规则：

• A的计数器：前9个A累计1-9；第10个A（索引10）继续累计到10；接下来3个A（索引13-15）累计到11-13；最后1个A（索引19）因为之前B连续出现3次，重置为1
• B的计数器：第1个B（索引9）因为之前A连续出现9次，重置为1；接下来2个B（索引11-12）累计到2-3；之后3个B（索引16-18）因为之前A连续出现3次，重置为1-3；最后2个B（索引20-21）继续累计到4-5

内容的提问来源于stack exchange，提问作者mali