如何将.fillna()与基于条件的字典一起使用

2024-05-16 19:26:20 发布

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我正在做一些房地产数据清理,遇到了这个新手的问题,令人惊讶的是,似乎我不能自己解决。你知道吗

我有一个数据帧,在lat和lon列中有nan值。输入给定邻域的lat和lon的平均值,我可以算出几乎正确的值。你知道吗

这是一个例子,实际的DF有超过20k行。你知道吗

    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,但是我不知道如何为fillna分配一个条件,所以它根据邻域lat和lon的平均值来填充lat和lon。你知道吗

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

谢谢你的帮助。你知道吗


Tags: 数据dfnanmean邻域monte平均值lon
1条回答
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1楼 · 发布于 2024-05-16 19:26:20

再次回答我自己的问题。。。你知道吗

在这个答案的帮助下,我找到了解决问题的正确代码: answer

代码:

创建词典:

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))

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