Python 如何根据条件将.fillna（）与字典一起使用

我正在做一些房地产数据清理，遇到了这个新手问题，令人惊讶的是，我自己似乎无法解决这个问题

我有一个数据帧，在lat和lon列中有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DF有超过20k行

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

我为每个社区制作了两本带有lat和lon意思的字典，代码如下：

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})

这是其中一条格言的一部分：

 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}

然后我试着用那些字典来填充lat和lon，但我不明白如何为fillna赋值，所以它根据邻域lat和lon的平均值来填充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

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谢谢你的帮助。

再次回答我自己的问题

借助以下答案，我找到了解决问题的正确代码：

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

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

使用字典填充nan值：

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

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

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你想用每个街区的平均值来填充nan，对吗？如果是这种情况，请增加数据量，使每个邻域在数据中不止一次。实际数据集包含超过20k行。这是一个示例，可能是的重复项，但在这种情况下，它们映射的是整个列，而不仅仅是nan值