基於條件語句在pandas數據框中生成新列

Question

我有 2 個熊貓數據框。 第一個包含站點的經緯度信息，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

第二個包含列“多邊形”（勻稱多邊形格式）並有 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, ...

我的目標是：

1. 檢查點（帶有她的坐標的站）是否在多邊形中（這沒問題）
2. 計算一個多邊形中的站點數量（這是問題所在）

我該怎么辦：

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，最后我想再多列一列“多邊形中的站數”，例如：

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

請問有什么想法嗎？ 非常感謝！

Answer 1

首先定義一個函數，該函數在給定的行上迭代站列表並計算該站是否包含在多邊形內。 然后在多邊形列表 DataFrame 的每一行上應用這個函數。

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')