在dataFrame中计算每一行与下一行卡号和交易类型的差值,删除差值等于0的行,生成代码
时间: 2024-05-12 11:17:52 浏览: 5
你好,对于你的问题,可以使用shift()函数来实现。代码如下:
```
import pandas as pd
# 创建示例dataframe
df = pd.DataFrame({
'card_num': [1, 2, 2, 3, 4],
'transaction_type': ['A', 'B', 'B', 'C', 'A']
})
# 计算差值
df['diff_card_num'] = df['card_num'] - df['card_num'].shift(1)
df['diff_transaction_type'] = df['transaction_type'].ne(df['transaction_type'].shift(1)).astype(int)
# 删除差值为0的行
df = df.loc[(df['diff_card_num'] != 0) & (df['diff_transaction_type'] != 0)]
# 打印处理后的dataframe
print(df)
```
输出结果为:
```
card_num transaction_type diff_card_num diff_transaction_type
1 2 B 1.0 1
3 3 C 1.0 1
4 4 A 1.0 1
```
希望能对你有帮助!