将邮政编码 API 调用应用于数据框中的每一行

Question

在下面的代码块中，我有一个数据框 geo，我想对其进行迭代以获得 geo 中每个英国邮政编码的东距、北距、经度和纬度。我写了一个函数来调用 API 和另一个函数来调用 return 这四个变量。

我已经用邮政编码测试了 get_data 调用以证明它有效（这是 public API 任何人都可以使用的）：

import requests 
import pandas as pd

geo = spark.table('property_address').toPandas()


def call_api(url: str) -> dict:
  postcode_response =requests.get(url)
  return postcode_response.json()

def get_data(postcode):

  url = f"http://api.getthedata.com/postcode/{postcode}"
  
  req = r.get(url)
  

  results = req.json()['data']
  easting = results['easting']
  northing = results['northing']
  latitude = results['latitude']
  longitude = results ['longitude']
  
  return easting ,northing,latitude, longitude

get_data('SW1A 1AA')

其中 return 个：

Out[108]: (529090, 179645, '51.501009', '-0.141588')

我想要做的是运行将 geo 和 return 中的每一行作为数据集。我的研究让我找到了 apply，我的尝试基于 this guide。

我试图在 geo 中传递一个名为 property_postcode 的列，并将每一行迭代到 return 值，这是我的尝试：

def get_columns(row):
  column_name = 'property_postcode'
  api_param = row[column_name]
  easting,northing,latitude,longitude = get_data(api_param)
  row['east'] = easting
  row['north'] = northing
  row['lat'] = latitude
  row['long'] = longitude
  return row

geo= geo.apply(get_columns, axis=1)

display(geo)

我得到的错误是

`JSONDecodeError: Expecting value: line 1 column 1 (char 0)`

并没有告诉我很多。寻找 assistance\pointers.

Answer 1

而不是尝试在函数中设置东、北、纬度和经度列的值 return 它们来自函数。

from numpy import result_type
import requests
import pandas as pd

# geo = spark.table('property_address').toPandas()


def call_api(url: str) -> dict:
    postcode_response = requests.get(url)
    return postcode_response.json()


def get_data(postcode):
    url = f"http://api.getthedata.com/postcode/{postcode}"
    req = requests.get(url)

    if req.json()["status"] == "match":
        results = req.json()["data"]
        easting = results.get("easting")
        northing = results.get("northing")
        latitude = results.get("latitude")
        longitude = results.get("longitude")
    else:
        easting = None
        northing = None
        latitude = None
        longitude = None

    return easting, northing, latitude, longitude


def get_columns(code):
    api_param = code
    return get_data(api_param)


df = pd.DataFrame(
    {
        "property_postcode": [
            "BE21 6NZ",
            "SW1A 1AA",
            "W1A 1AA",
            "DE21",
            "B31",
            "ST16 2NY",
            "S65 1EN",
        ]
    }
)

df[["east", "north", "lat", "long"]] = df.apply(
    lambda row: get_columns(row["property_postcode"]), axis=1, result_type="expand"
)

print(df)

property_postcode	east	north	lat	long
BE21 6NZ	NaN	NaN	None	None
SW1A 1AA	529090	179645	51.501009	-0.141588
W1A 1AA	528887	181593	51.518561	-0.143799
DE21	NaN	NaN	None	None
B31	NaN	NaN	None	None
ST16 2NY	391913	323540	52.809346	-2.121413
S65 1EN	444830	394082	53.44163	-1.326573

将邮政编码 API 调用应用于数据框中的每一行

Apply postcode API call to each row in dataframe

python

pyspark

databricks