Python 熊猫：基于索引合并数据帧和系列

我创建了以下数据框

priceearning\u byyear

，但仍然不完整：

之后，我想将以下系列

price

插入/合并到上述df中：

这个系列有一行（2021）太多了，我想在插入上面的df时删除它

我的尝试是：

priceearning_byyear.merge(price.to_frame(), left_index=True)

但我犯了以下错误：

MergeError: Must pass right_on or right_index=True

如果我使用

right\u index=True

，我将得到：

MergeError: Must pass left_on or left_index=True

---

我不明白，为什么它不起作用。感谢您提供的任何指针：-）

您需要在

priceearning\u byyear

中指定要合并的列

data1 = {"year":[x for x in range(2016, 2020)], "eps":[x for x in range(2,12,2)]}
data2 = pd.Series(data = {2016: 1, 2017: 2, 2018: 3, 2019: 4, 2020: 5, 2021: 6})

df1 = pd.DataFrame(data1)
df1.merge(data2.to_frame(name='price'), how='left', left_on="year", right_index=True)

---

您需要在

priceearning\u byyear

中指定要合并的列

data1 = {"year":[x for x in range(2016, 2020)], "eps":[x for x in range(2,12,2)]}
data2 = pd.Series(data = {2016: 1, 2017: 2, 2018: 3, 2019: 4, 2020: 5, 2021: 6})

df1 = pd.DataFrame(data1)
df1.merge(data2.to_frame(name='price'), how='left', left_on="year", right_index=True)

---

你就快到了。只需同时通过

left_index

和

right_index

即可获得如下所示的效果

# Reproduce your data
import pandas as pd
priceearning_byyear = pd.DataFrame(dict(year=[2016,2017,2018,2019,2020], eps=[2.09,2.32,3.00,2.99,3.31])).set_index('year')
price = pd.Series([28.95,42.31,39.44,73.41,132.69,119.99], index=[2016,2017,2018,2019,2020,2021])
price.name = 'Close'
price.index.name='year'

# Merge priceearning_byyear and price by using their indexes
priceearning_byyear.merge(price,left_index=True,right_index=True)

---

你就快到了。只需同时通过

left_index

和

right_index

即可获得如下所示的效果

# Reproduce your data
import pandas as pd
priceearning_byyear = pd.DataFrame(dict(year=[2016,2017,2018,2019,2020], eps=[2.09,2.32,3.00,2.99,3.31])).set_index('year')
price = pd.Series([28.95,42.31,39.44,73.41,132.69,119.99], index=[2016,2017,2018,2019,2020,2021])
price.name = 'Close'
price.index.name='year'

# Merge priceearning_byyear and price by using their indexes
priceearning_byyear.merge(price,left_index=True,right_index=True)

---

这篇文章可能会有帮助这篇文章也可能会帮助你的代码工作。谢谢：-）您的代码也可以工作。谢谢：-）