Python 3.x 将函数聚合到groupby

我想写一个查询，计算每个太阳系中相应恒星大于太阳的行星数量（即它们的半径大于1）

您的查询应该返回恒星的半径和行星数，只显示行星数多于一的行。结果应根据恒星半径按降序排序

import pandas as pd
import numpy as np

df_star = pd.read_csv("stars.csv",names=["kepler_id","t_eff","radius"])
df_planet = pd.read_csv("planets.csv",names=["kepler_id","koi_name","kepler_name","status","period","radius","t_eq"])

df_star[df_star["radius"]>1].merge(df_planet,on="kepler_id").groupby("radius_x",sort=True).count()

我有以下结果 [！[在此处输入图像描述][1][1]

但我只想返回大于1的计数

我也试过了

agg({'kepler_id': lambda x: len(x>1)})

来自行星的数据如下：（样本）

对于星星（大文件的示例）

---

请以可复制的格式发布您的输入数据。您能用文字解释一下您想要做什么吗？为什么要按

radius\x

分组？两颗半径相同的恒星不算是两颗独立的恒星吗？不算。恒星的半径相对于太阳，行星的半径相对于地球。

10666592,K00002.01,Kepler-2 b,CONFIRMED,2.204735365,16.39,2025
6922244,K00010.01,Kepler-8 b,CONFIRMED,3.522498573,14.83,1521
11904151,K00072.01,Kepler-10 b,CONFIRMED,0.837491331,1.45,1968
10187017,K00082.04,Kepler-102 c,CONFIRMED,7.07136076,0.58,723
10187017,K00082.05,Kepler-102 b,CONFIRMED,5.28695437,0.49,797
10984090,K00112.02,Kepler-466 c,CONFIRMED,3.709213846,1.24,1236
9579641,K00115.01,Kepler-105 b,CONFIRMED,5.41220713,3.28,1306
9579641,K00115.02,Kepler-105 c,CONFIRMED,7.12594591,1.88,1191
9579641,K00115.03,,CANDIDATE,3.4358789,0.65,1519
8395660,K00116.01,Kepler-106 c,CONFIRMED,13.57076622,2.35,796
8395660,K00116.02,Kepler-106 e,CONFIRMED,43.84444353,2.58,538
8395660,K00116.03,Kepler-106 b,CONFIRMED,6.16491696,0.85,1035
8395660,K00116.04,Kepler-106 d,CONFIRMED,23.9802348,0.99,658
10875245,K00117.02,Kepler-107 c,CONFIRMED,4.90143807,1.84,1263
  [1]: https://i.stack.imgur.com/BO6Cw.png

2713049,5996,0.956
3114167,5666,0.677
3115833,5995,0.847
3246984,5735,0.973
3342970,6167,1.064
3351888,5717,1.057
3453214,5733,0.77
3641726,5349,0.82
3832474,5485,0.867
3935914,5934,0.893
3940418,5170,0.807
4049131,4905,0.761
4139816,3887,0.48
4275191,5557,0.781
4476123,5413,0.751
5358241,6079,0.945
5358624,5071,0.788
5456651,4980,0.734
6862328,5796,0.871
6922244,6225,1.451