当一组列表只分配给一个变量时，如何将多个列表连接到一个列表中?
大多数在线建议显示两个或多个变量连接在一起，但我的只是一个变量分配给许多列表。我尝试了一个嵌套的For-Loop，但是导致了重复和不连贯的列表。还尝试了扩展和追加函数，但没有成功。也许我应该用数据框架来解决这个问题?
任何帮助都是非常感激的。如果你有问题，请提出来。 实际代码:
from bs4 import BeautifulSoup as bs
import requests
import re
from time import sleep
from random import randint
def price():
baseURL='https://www.apartmentlist.com/ca/redwood-city'
r=requests.get(baseURL)
soup=bs(r.content,'html.parser')
block=soup.find_all('div',class_='css-1u6cvl9 e1k7pw6k0')
sleep(randint(2,10))
for properties in block:
priceBlock=properties.find_all('div',class_="css-q23zey e131nafx0")
price=[price.text for price in priceBlock]
strPrice=''.join(price) #Change from list to string type
removed=r'[$]' #Select and remove $ characters
removed2=r'Ask' #Select and remove Ask
removed3=r'[,]' #Select and remove comma
modPrice=re.sub(removed,' ',strPrice) #Substitute $ for '_'
modPrice2=re.sub(removed2,' 0',modPrice) #Substitute Ask for _0
modPrice3=re.sub(removed3,'',modPrice2) #Eliminate space within price
segments=modPrice3.split() #Change string with updates into list, remain clustered
for inserts in segments:
newPrice=[inserts] #Returns values from string to list by brackets.
print(newPrice)
price()
实际输出:
#After executing the program
['2157']
['2805']
['0']
['1875']
['2800']
['2265']
['2735']
['3985']
...
...
尝试:
['2157', '2805', '0', '2800',...] # all the while assigned to a single variable.
再次感谢您的帮助。
您的代码中的问题是，在段中插入的"循环只接受每个价格，将其放入自己的列表中，然后输出仅包含1个内容的列表。因此，您需要将所有价格添加到同一列表中，然后在循环后输出它。
在你的情况下，你可以使用像这样的列表推导来实现你想要的:
from bs4 import BeautifulSoup as bs
import requests
import re
from time import sleep
from random import randint
def price():
baseURL='https://www.apartmentlist.com/ca/redwood-city'
r=requests.get(baseURL)
soup=bs(r.content,'html.parser')
block=soup.find_all('div',class_='css-1u6cvl9 e1k7pw6k0')
sleep(randint(2,10))
result = []
for properties in block:
priceBlock=properties.find_all('div',class_="css-q23zey e131nafx0")
price=[price.text for price in priceBlock]
strPrice=''.join(price) #Change from list to string type
removed=r'[$]' #Select and remove $ characters
removed2=r'Ask' #Select and remove Ask
removed3=r'[,]' #Select and remove comma
modPrice=re.sub(removed,' ',strPrice) #Substitute $ for '_'
modPrice2=re.sub(removed2,' 0',modPrice) #Substitute Ask for _0
modPrice3=re.sub(removed3,'',modPrice2) #Eliminate space within price
segments=modPrice3.split() #Change string with updates into list, remain clustered
result += [insert for insert in segments]
print(result)
price()
(希望我理解这个问题)
如果每个子列表都是一个变量，则可以执行以下操作之一将它们转换为单个列表:
a = ['2157']
b = ['2805']
c = ['0']
d = ['1875']
e = ['2800']
f = ['2265']
g = ['2735']
h = ['3985']
#Pythonic Way
test = [i[0] for i in [a, b, c, d, e, f, g, h]]
print(test)
#Detailed Way
check = []
for i in a,b,c,d,e,f,g,h:
check.append(i[0])
print(check)
如果你的函数创建列表，那么你只需要修改for循环来引用你的函数:
#Pythonic Way
test = [i[0] for i in YOUR_FUNCTION()]
print(test)
#Detailed Way
check = []
for i in YOUR_FUNCTION():
check.append(i[0])
print(check)
如何将多个列表连接到一个列表中，当列表的集合只分配给一个变量时?
在Python中，通常使用列表推导式或itertools.chain来扁平化列表的列表。
from itertools import chain
prices = [
['2157'],
['2805'],
['0'],
['1875'],
['2800'],
['2265'],
['2735'],
['3985'],
]
# list comprehension
[x for row in prices for x in row]
>>> ['2157', '2805', '0', '1875', '2800', '2265', '2735', '3985']
# itertools.chain will return a generator like object
chain.from_iterable(prices)
>>> <itertools.chain at 0x7f01573076a0>
# if you want a list back call list
list(chain.from_iterable(prices))
>>> ['2157', '2805', '0', '1875', '2800', '2265', '2735', '3985']
对于上面的代码，price函数只打印输出而不返回对象。您可以让函数创建一个空列表，并在每次循环遍历属性时向列表中添加内容。然后返回列表。
def price():
# web scrape code
new_price = []
for properties in block:
# processing code
new_price += [x for x in segments]
return chain.from_iterable(new_prices)