在 pandas 数据框上使用布尔过滤器时出现 KeyError

Question

当一个数据框中的日期时间对象在另一个数据框中的日期时间对象范围内时，尝试合并两个数据框。

继续收到：KeyError: 'cannot use a single bool to index into setitem' 在我发布的第二个块的这一行代码上。

gametaxidf.loc[arrivemask, 'relevant'] = 1

我假设它也会在下一行使用类似的命令发生。

这是给我带来麻烦的部分：

with open('/Users/benjaminprice/Desktop/TaxiCombined/Data/combinedtaxifiltered.csv', 'w') as csvfile: 
    fieldnames1 = ['index','pickup_datetime', 'dropoff_datetime', 'pickup_long', 'pickup_lat','dropoff_long','dropoff_lat','passenger_count','trip_distance','fare_amount','tip_amount','total_amount','stadium_code'] 
    writer = csv.DictWriter(csvfile, fieldnames=fieldnames1) 
    writer.writeheader()

for index, row in baseballdf.iterrows(): 
    gametimestart = row['Start.Time'] 
    gametimeend = row['End.Time'] 
    arrivemin = gametimestart - datetime.timedelta(minutes=120) 
    arrivemax = gametimeend - datetime.timedelta(minutes = 30) 
    departmin = gametimeend - datetime.timedelta(minutes = 60) 
    departmax = gametimeend + datetime.timedelta(minutes = 90)

    gametaxidf = combineddf[combineddf.DATE==row.DATE]
    gametaxidf['relevant']=0

    for index, row in gametaxidf.iterrows():
        arrivemask = (arrivemin < row['dropoff_datetime']) and (row['dropoff_datetime'] < arrivemax)
        departmask = (departmin < row['pickup_datetime']) and (row['pickup_datetime'] < departmax) 
        gametaxidf.loc[arrivemask, 'relevant'] = 1
        gametaxidf.loc[departmask, 'relevant'] = 1

        with open('/Users/benjaminprice/Desktop/TaxiCombined/Data/combinedtaxifiltered.csv','a') as combinedtaxi:
            gametaxidf.to_csv(combinedtaxi,header=None)
    print(str(index) + "done")

Gametaxidf.head(5):

   index     pickup_datetime    dropoff_datetime  pickup_long  pickup_lat  \
0    195 2014-04-01 00:08:13 2014-04-01 00:15:32   -73.922218   40.827557   
1    344 2014-04-01 00:16:30 2014-04-01 00:20:38   -73.846046   40.754566   
2    558 2014-04-01 00:28:59 2014-04-01 00:36:36   -73.921692   40.831394   
3    744 2014-04-01 00:42:00 2014-04-01 00:49:46   -73.938080   40.804646   
4    776 2014-04-01 00:43:54 2014-04-01 00:53:22   -73.952652   40.810577   

   dropoff_long  dropoff_lat  passenger_count  trip_distance  fare_amount  \
0    -73.900620    40.856174                1           2.30          9.0   
1    -73.890259    40.753246                1           0.56          4.5   
2    -73.942719    40.823257                1           1.53          7.0   
3    -73.928490    40.830433                1           2.96         11.0   
4    -73.924332    40.827320                1           2.28         10.5   

   tip_amount  total_amount  stadium_code       DATE  relevant  
0           0          10.0           1.1 2014-04-01         0  
1           0           5.5           2.1 2014-04-01         0  
2           0           8.0           1.1 2014-04-01         0  
3           0          12.0           1.0 2014-04-01         0  
4           0          11.5           1.0 2014-04-01         0

同时收到此警告：试图在 DataFrame 的切片副本上设置一个值。

Try using .loc[row_indexer,col_indexer] = value instead

但这让我可以继续……任何帮助都会很棒。

Answer 1

这里

gametaxidf.loc[arrivemask, 'relevant'] = 1

您正在尝试通过 .loc 运算符设置数据帧值。 Pandas docs for selecting rows 说：

.loc is primarily label based, but may also be used with a boolean array. .loc will raise KeyError when the items are not found. Allowed inputs are:

A single label, e.g. 5 or 'a', (note that 5 is interpreted as a label of the index. This use is not an integer position along the index)

A list or array of labels ['a', 'b', 'c']

A slice object with labels 'a':'f', (note that contrary to usual python slices, both the start and the stop are included!)

A boolean array

您正在尝试使用最后一种输入方式，但是

arrivemask = (arrivemin < row['dropoff_datetime']) and 
    (row['dropoff_datetime'] < arrivemax)

是标量布尔值，不是数组。

您无需遍历数据框。 Pandas 为你做。只需使用：

gametaxidf.loc[
   (arrivemin < gametaxidf['dropoff_datetime'])
   &
   (gametaxidf['dropoff_datetime'] < arrivemax)
   , 'relevant'] = 1

在 pandas 数据框上使用布尔过滤器时出现 KeyError

KeyError when using boolean filter on pandas data frame

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