combined_df = pd.concat([data1, data2, data3, data4, data5, data6, data7], axis=0)

Try with out = df.groupby('A',as_index=False).agg({'B':','.join,'C':'first'}) A B C0 a d,f,e i1 c g j Update out = df.groupby('A',as_index=False).agg({'B':lambda x : ','.join(x.dropna().drop_duplicates()),'C':lambda x : ','.join(x.dropna().drop_duplicates())})out A B C0 a d,f,e i1 c g j

最簡單的事情可能就是定義一個方法： import dataclassesimport jax.numpy as jnp@dataclasses.dataclassclass Metric: score1: jnp.ndarray score2: jnp.ndarray score3: jnp.ndarray def concatenate(self, other): return Metric( jnp.concatenate((self.score1, other.score1)), jnp.concatenate((self.score2, other.score2)), jnp.concatenate((self.score3, other.score3)), ) 然后做a.concatenate(b)。您也可以調用方法__add__，這樣就可以只使用a + b。這更整潔，但可能與element-wise加法混淆。

您可以在pro1上groupby并從pro2創建列表，最后轉換為字典： out = df.groupby('pro1')['pro2'].agg(list).to_dict()

可能有許多選擇之一是 from functools import reduceimport pandas as pdidx = ["Year", "Month"]new_df = reduce(pd.DataFrame.join, (i.set_index(idx) for i in dataframes)).reset_index() or reduce(lambda x, y: pd.merge(x, y, how="outer", on=["Year", "Month"]), dataframes)

用groupby/cumcount創建一個temp列t，然后將該列用于合并。 merged_df = ( df1.assign(t=df1.groupby('Date').cumcount()) .merge( df2.assign(t=df2.groupby('Quarter').cumcount()), left_on=['Date', 't'], right_on=['Quarter', 't'], how='left') .drop('t', 1)) OUTPUT: ID Date Quarter State0 a 1/1/2022 1/1/2022 ny1 a 1/1/2022 NaN NaN2 b 2/1/2022 NaN NaN3 b 2/1/2022 NaN NaN4 b 2/1/2022 NaN NaN

您需要執行交叉聯接。以下是我的建議： # Creating a dummy column to perform a joindf1["Dummy"] = 1df2["Dummy"] = 1# to obtain the cross join we will merge on the dummy column and drop it.df_desired= pd.merge(df1, df2, on ='Dummy').drop("Dummy", 1)

使用append，然后使用drop_duplicates data1.drop('Low',1).append([data2, data3], ignore_index=True).drop_duplicates()

更新：您的新示例數據仍然可以很好地合并，所以您的實際列可能有虛假的空格。在合并之前嘗試strip()它們： df1.Date = df1.Date.str.strip()df2.Date = df2.Date.str.strip()df1.Name = df1.Name.str.strip()df2.Name = df2.Name.str.strip()merged = df1.merge(df2, on=['ID', 'Date', 'Name'])# ID Date Name Volume Up Volume Down# 0 1 2019-02-01 to 2019-03-15 Call 50.0 66.0# 1 1 2019-02-01 to 2019-03-15 Email 60.5 50.0# 2 1 2019-02-01 to 2019-03-15 Radio 20.0 40.0# 3 2 2019-02-01 to 2019-03-15 Call 5.5 10.0# 4 2 2019-02-01 to 2019-03-15 Email 6.4 12.2# 5 2 2019-02-01 to 2019-03-15 Radio 15.0 7.0 您可以給merge()一個列列表來合并on： out = df1.merge(df2, on=['ID', 'Date', 'N

import pandas as pdfrom pandas import Series, DataFramedf1 = pd.DataFrame({'values': [1, 2, 3]}, index=['a', 'b', 'c'])df2 = pd.DataFrame({'values': [4, 5, 6]}, index=['c', 'd', 'e'])df_new = pd.concat([df1, df2], axis=0)print(df_new)df_new1 = df_new.groupby(df_new.index).first()print(df_new1)df_new2 = df_new.groupby(df_new.index).last()print(df_new2) valuesa 1b 2c 3c 4d 5e 6 valuesa 1b 2c 3d 5e 6 valuesa 1b 2c 4d 5e 6

對于Id和Flower列的前4個字符的substr合并，請嘗試如下操作： import pandas as pddf1 = pd.DataFrame({'Flower': ["Jasmine", "Lotus", "BudRose", "Lily"], 'Id': ["1023", "3405", "4409", "2457"], 'City': ["Hawai", "St Jose", "Miami", "Washington"]})df2 = pd.DataFrame({'Flower': ["Jasmine", "Lily", "Lily"], 'Id': ["1023LD", "2457MH", "3346HK", ], 'City': ["Hawai", "Washington", "Washington"]})def get_substr(x): return x['Id'].str[:4]merged_df = df1.assign(merge_with_x=get_substr) \ .merge(df2.assign(merge_with_x=get_substr), on=['Flower', 'City', 'merge_with_x'], suffixes=['_x', '/z']) \ .filter(regex=r'Flower|City|.*/z')#

您可以嘗試使用pd.concat，然后使用groupby： main_df = pd.concat([main_df,df2]).groupby('Type', sort=False).agg({'Count': sum}).reset_index() OUTPUT: Type Count0 Album 211 Book 42 Person 73 Film 4

Lets say df=表1 genre total0 action 4501 adventure 2002 animation 3003 crime 10 df2=表2 genre total0 action 4501 adventure 2002 animation 300 這些可以通過以下方式完成： f = [df, df2]result = pd.concat(f)new_df = result.groupby('genre')['total'].sum().reset_index() output genre total0 action 9001 adventure 4002 animation 6003 crime 10

您可以下載tickers列表： tickers = ['^VIX', 'GLD', '^TNX', '^GSPC']df = yf.download(tickers, '2008-01-01', date.today(), auto_adjust=True)['Close']print(df)# Output: GLD ^GSPC ^TNX ^VIXDate 2007-12-31 82.459999 1468.359985 4.035 22.5000002008-01-02 84.860001 1447.160034 3.901 23.1700002008-01-03 85.570000 1447.160034 3.901 22.4900002008-01-04 85.129997 1411.630005 3.854 23.9400012008-01-07 84.769997 1416.180054 3.839 23.790001... ... ... ... ...2021-12-02 165.240005 4577.100098 1.448 27.9500012021-12-03 166.630005 4538.430176 1.343 30.6700002021-12-06 166.220001 4591.669922 1.434 27.1800002021-12-07 16

Try: second["Code"] = second["Code"].str.split(",")second = second.explode("Code")first["Code"] = first["Code"].astype(str)second["Code"] = second["Code"].astype(str)third = pd.merge(first, second, on="Code")third["Value"] = third.apply(lambda x: f"{x['Value']}({x['Code']})", axis=1)print( third.groupby("Product id", as_index=False) .agg({"Code": ",".join, "Value": ",".join}) .rename(columns={"Value": "Required Field"})) Prints: Product id Code Required Field0 A 1 Apple(1)1 B 2 Orange(2)2 C 1 Apple(1)3 D 2,4 Orange(2),Graps(4)4 E 3 Bannana(3)