根据条件语句在熊猫数据框中生成新列

Question

我有 2 个熊猫数据框。第一个包含站点的经纬度信息，只有 3 行：

    stat_id    stat_lon     stat_lat
0   db_695203   9.444328    54.787590
1   db_699007   9.438629    54.789577
2   db_695442   9.445865    54.786215

第二个包含列'Polygon'（匀称的多边形格式）并且有20行：

0     POLYGON ((9.444721146384639 54.78805404001241,...
1     POLYGON ((9.429828147969117 54.79003403977831,... 
2     POLYGON ((9.429153147576411 54.78516304109078,...
.......................................................
18    POLYGON ((9.417355147148637 54.79108504035977,...
19    POLYGON ((9.44272277037326 54.79218198146992, ...

我的目标是：

1. 检查点（带有她的坐标的站）是否在多边形中（这个没问题）
2. 计算一个多边形中的站数（这就是问题所在）

我该怎么办：

for j in range(len(piece_clean_data)): #it's a df which contains polygons
P = shapely.wkt.loads(piece_clean_data.iloc[j,87]) #i convert string to Polygon
for i in range(len(three_stations)): #df with 3 stations
    p1 = Point(three_stations.iloc[i,1], three_stations.iloc[i,2]) #station coordinates
    st = P.contains(p1) #the answer is "True/False" - here i check, whether the point is in polygon or not
    if st == 'True': #and here I don't have any idea.

所以，最后我想多写一列“多边形中的站点数”，例如：

    0     POLYGON ((9.444721146384639 54.78805404001241,...   0
    1     POLYGON ((9.429828147969117 54.79003403977831,...   0 
    2     POLYGON ((9.429153147576411 54.78516304109078,...   1

请问有什么想法吗？提前非常感谢！

Answer 1

首先定义一个函数，给定一行迭代站点列表并计算站点是否包含在多边形内。然后对多边形列表DataFrame的每一行应用这个函数。

def num_stations(polygon):
  """Count the number of stations that are within a given polygon."""
  num = 0
  for _, station in three_stations.iterrows():
    p = Point(station['stat_lon'], station['stat_lat'])
    if polygon.contains(p):
      num += 1
  return num

piece_clean_data['station_counts'] = piece_clean_data.apply(num_stations, axis='columns')