Question

我有一些从csv中查找的代码，然后如果csv中不存在它，则从google地图中检索它。我有100,000多条记录，大约需要2个小时。关于如何加快速度的任何想法？谢谢！

from csv import DictReader
import codecs

def find_school(high_school, city, state):
    types_of_encoding = ["utf8"]
    for encoding_type in types_of_encoding:
        with codecs.open('C:/high_schools.csv', encoding=encoding_type, errors='replace') as csvfile:
            reader = DictReader(csvfile)
            for row in reader:
            #checks the csv file and sees if the high school already exists
                if (row['high_school'] == high_school.upper() and
                    row['city'] == city.upper() and
                    row['state'] == state.upper()):
                    return dict(row)['zipcode'],dict(row)['latitude'],dict(row)['longitude'],dict(row)['place_id']
                else:
                    #hits Google Maps api
#executes
df['zip'],df['latitude'], df['longitude'], df['place_id'] = zip(*df.apply(lambda row: find_school(row['high_school'].strip(), row['City'].strip(), row['State'].strip()), axis=1))

CSV文件片段

high_school,city,state,address,zipcode,latitude,longitude,place_id,country,location_type
GEORGIA MILITARY COLLEGE,MILLEDGEVILLE,GA,"201 E GREENE ST, MILLEDGEVILLE, GA 31061, USA",31061,33.0789184,-83.2235169,ChIJv0wUz97H9ogRwuKm_HC-lu8,USA,UNIVERSITY
BOWIE,BOWIE,MD,"15200 ANNAPOLIS RD, BOWIE, MD 20715, USA",20715,38.9780387,-76.7435378,ChIJRWh2C1fpt4kR6XFWnAm5yAE,USA,SCHOOL
EVERGLADES,MIRAMAR,FL,"17100 SW 48TH CT, MIRAMAR, FL 33027, USA",33027,25.9696495,-80.3737813,ChIJQfmM_I6j2YgR1Hdq0CC4apo,USA,SCHOOL

Answer 1

每次检查都没有必要读取文件。只需将文件加载到内存中一次，然后使用元组键的一部分将您感兴趣的字段创建一个新字典即可。

import csv

lookup_dict = {}
with open('C:/Users/Josh/Desktop/test.csv') as infile:
    reader = csv.DictReader(infile)
    for row in reader:
        lookup_dict[(row['high_school'].lower(), row['city'].lower(),
                    row['state'].lower())] = row

现在，您只需检查要测试的值是否已经是lookup_dict中的键。如果不是，则查询Google Maps。

由于您的编辑显示您正在使用此方法apply到数据框，因此应在函数外部计算lookup_dict并将其作为参数传递。这样，该文件仍然只能读取一次。

lookup_dict = {}
with open('C:/Users/Josh/Desktop/test.csv') as infile:
    reader = csv.DictReader(infile)
    for row in reader:
        lookup_dict[(row['high_school'].lower(), row['city'].lower(),
                    row['state'].lower())] = row

def find_school(high_school, city, state, lookup_dict):
    result = lookup_dict.get((high_school.lower(), city.lower(), state.lower()))
    if result:
        return result
    else:
        # Google query
        pass

a = find_school('georgia military college', 'milledgeville', 'ga', lookup_dict)
#df['zip'],df['latitude'], df['longitude'], df['place_id'] = zip(*df.apply(lambda row: find_school(row['high_school'].strip(), row['City'].strip(), row['State'].strip()), axis=1))

在CSV中检查值的更快方法？

1 个答案: