如何根据条件将.fillna()与字典结合使用

时间:2019-09-28 17:25:33

标签: python pandas numpy dataframe

我正在做一些房地产数据清理工作,遇到了这个新手问题,令人惊讶的是我自己无法解决。

我有这个数据帧,在lat和lon列中具有nan值。我可以算出输入给定邻域的lat和lon平均值的几乎正确的值。

这是一个示例,实际的DF有超过2万行。

    lat   lon    neighborhood
   -34.62 -58.50 Monte Castro
   -34.63 -58.36 Boca
    nan   nan    San Telmo

我用以下代码制作了两个带有lat和lon意思的字典,用于每个邻域:

neighborhood_lat = []
neighborhood_lon = []
for neighborhood in df['l3'].unique():
    lat = df[((df['l3']==neighborhood) & (df['lat'].notnull()))].mean().lat
    lon = df[((df['l3']==neighborhood) & (df['lon'].notnull()))].mean().lon
    neighborhood_lat.append({neighborhood: lat})
    neighborhood_lon.append({neighborhood: lon})

这是其中一项命令的一部分:

 neighborhood_lat 
 [{'Mataderos': -34.65278757721805},
 {'Saavedra': -34.551813882357166},
 {nan: nan},
 {'Boca': -34.63204552441155},
 {'Boedo': -34.62695442446412},
 {'Abasto': -34.603728937455315},
 {'Flores': -34.62757516061659},
 {'Nuñez': -34.54843158034983},
 {'Retiro': -34.595564030955934},
 {'Almagro': -34.60692879236826},
 {'Palermo': -34.58274909271148},
 {'Belgrano': -34.56304387233704},
 {'Recoleta': -34.592081482406854},
 {'Balvanera': -34.608665174550694},
 {'Caballito': -34.61749059613885}

然后我试图用这些词典填充lat和lon,但是我不明白如何为fillna设置条件,以便根据邻居lat和lon的意思填充lat和lon。

预期结果

    lat                         lon                       neighborhood
   -34.62                      -58.50                     Monte Castro
   -34.63                      -58.36                     Boca
    (mean lat of neighborhood) (mean lon of neighborhood) San Telmo

感谢您的帮助。

1 个答案:

答案 0 :(得分:0)

再次回答我自己的问题...

我在此答案的帮助下找到了解决该问题的正确代码: answer

代码:

创建字典:

neighborhood_lat = {}
neighborhood_lon = {}

for neighborhood in df['l3'].unique():
    neighborhood_lat[neighborhood] = df[((df['l3']==neighborhood) & (df['lat'].notnull()))].mean().lat
    neighborhood_lon[neighborhood] = df[((df['l3']==neighborhood) & (df['lon'].notnull()))].mean().lon

用字典填充nan值:

df['lat'] = df['lat'].fillna(df['l3'].map(neighborhood_lat))
df['lon'] = df['lon'].fillna(df['l3'].map(neighborhood_lon))