Python 3.x 如何连接字符串并连接concurrent.futures生成的数据帧?
我有功能Python 3.x 如何连接字符串并连接concurrent.futures生成的数据帧?,python-3.x,pandas,dataframe,concatenation,concurrent.futures,Python 3.x,Pandas,Dataframe,Concatenation,Concurrent.futures,我有功能check_status,输入I并输出[a,b],其中a是字符串,b是数据帧。为了加速循环,我使用了concurrent.futures。您能解释一下如何连接所有字符串a并连接所有数据帧b import pandas as pd import concurrent.futures import requests session = requests.Session() headers = {'User-Agent': 'Mozilla/5.0 (X11; Ubuntu; Linux x8
check_status
,输入I
并输出[a,b]
,其中a
是字符串,b
是数据帧。为了加速循环,我使用了concurrent.futures
。您能解释一下如何连接所有字符串a
并连接所有数据帧b
import pandas as pd
import concurrent.futures
import requests
session = requests.Session()
headers = {'User-Agent': 'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:78.0) Gecko/20100101 Firefox/78.0'}
# Define a dictionary containing links
data = {'Link': ['https://www.collinsdictionary.com/dictionary/french-english/aimer',
'https://www.collinsdictionary.com/dictionary/english-french/graduate',
'https://www.collinsdictionary.com/dictionary/english-french/apple',
'https://www.collinsdictionary.com/dictionary/english-french/apricot'],
'Status': [0, 0, 0, 0]}
# Generate datafrane df
df = pd.DataFrame(data, columns = ['Link', 'Status'])
#Generate function to check status of a link
def check_status(i):
r = session.get(df.at[i, 'Link'], headers = headers)
if r.status_code == 200:
df.at[i, 'Status'] = 1
return 'ok', pandas.DataFrame({'LINK': ['abc'], 'CODE': [i]})
else:
return 'not ok'
with concurrent.futures.ThreadPoolExecutor(max_workers = 5) as exe:
results = [exe.submit(check_status, i) for i in [0, 1, 2, 3]]