我在 pandas 中有一個名為 names_and_places 的數(shù)據(jù)框字典，如下所示。姓名和地點(diǎn)：Alfred,,,Date,F_1,F_2,Key4/1/2020,1,4,NAN4/2/2020,2,5,NAN4/3/2020,3,6,"[USA,NY,NY, NY]"Brett,,,Date,F_1,F_2,Key4/1/2020,202,404,NAN4/2/2020,101,401,NAN4/3/2020,102,403,"[USA,CT, Fairfield, Stamford] "Claire,,,Date,F_1,F_2,Key4/1/2020,NAN,12,NAN4/2/2020,NAN,45,NAN4/3/2020,7,78,"[USA,CT, Fairfield, Darian] "Dane,,,Date,F_1,F_2,Key4/1/2020,4,17,NAN4/2/2020,5,18,NAN4/3/2020,7,19,"[USA,CT, Bridgeport, New Haven] "Edward,,,Date,F_1,F_2,Key4/1/2020,4,17,NAN4/2/2020,5,18,NAN 4/3/2020,7,19,"[USA,CT, Bridgeport, Milford] "（上面的文字或下面的圖片）
1 回答
qq_笑_17
TA貢獻(xiàn)1818條經(jīng)驗(yàn) 獲得超7個贊
這是一種更有效的方法。您首先從字典中構(gòu)建一個數(shù)據(jù)框，然后在該數(shù)據(jù)框上執(zhí)行實(shí)際工作。
single_df = pd.concat([df.assign(name = k) for k, df in names_and_places.items()])
single_df["Key"] = single_df.Key.replace("NAN", np.NaN)
single_df.dropna(inplace=True)
# Since the location is a string, we have to parse it.
location_df = pd.DataFrame(single_df.Key.str.replace(r"[\[\]]", "").str.split(",", expand=True))
location_df.columns = ["Country", "State", "County", "City"]
single_df = pd.concat([single_df, location_df], axis=1)
# this is where the actual query goes.
single_df[(single_df.Country == "USA") & (single_df.State == "CT")].name
輸出是：
2 Brett
2 Claire
2 Dane
2 Edward
Name: name, dtype: object
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