的Python：用熊猫GROUPBY，以减少数据帧

Question

的维度在我的数据框，我们称之为DF，我有一个看起来像

serial gps_dt lat long dist 
1  25Mar x1 y1 Nan 
1  26Mar x2 y2 0.01 
1  27Mar x3 y3 1.25 (assume this is the 5th occurrence < 160) 
2  24Mar x4 y5 Nan 
2  25Mar x5 y5 2.1 
2  26Mar x6 y6 1.01 
2  27Mar x7 y7 175.2 
2  28Mar x8 y8 179.3 (assume this is the 5th occurrence > 160) 

，这样下去的数据。我已经有一个系列，我们把它叫做check，告诉我是否serial[i] == serial[i+1]。我现在想要做的是当它们相等时，在条件hdist < 160下构造一个包含serial, gps_dt_first, gps_dt_last, avg_lat, avg_long的新数据帧，并且在此半径内至少有5次出现。如果hdist > 160，我想建造另一组当且仅当在未来5个事件是中第一个大于160

160例如，输出看起来是这样的：

serial gps_dt_first gps_dt_last avg_lat avg_long 
1  25Mar  27Mar  avg_x avg_y 
2  27Mar  28Mar  avg_x avg_y 

我我正在看熊猫的group by文档。该数据已经在SAS的serial, gps_dt订单中。我还需要做df.groupby(['serial', 'gps_dt'])吗？

一旦DF进行分组，如果需要的话，我的代码的思想是（更多的是伪代码大纲）：

if check == true and hdist < 160 and 5 or more occurrences (how to count the occurrences): 
    result['serial'] = df.serial (first in serial; how to extract) 
    result['gps_dt_first'] = df.gps_dt (first in gps_dt) 
    result['gps_dt_last'] = df.gps_dt (last in gps_dt) 
    result['avg_lat'] = df.lat.mean() (only for the subset of serial meeting criteria) 
    result['avg_long'] = df.long.mean() (same here) 
else if check == true and hdist > 160 and 5 or more occurrences; 
    do same as above 
else: 
    delete 

Answer 1

如果您已经阅读文档为groupby，你可以做什么以下部分解释：

1. Iterate over each element you got from groupby;
2. Perform one or more aggregate operations（包括应用链接操作，或根据列不同操作）;