如何在for循环中只向列表添加一次？

Question

代码正在工作并添加到列表中。然而，它是添加到每个列表三次，而不是一次。我想把列表中的项目追加一次，而不是三次。你知道吗

我试过检查范围，但只有一个。但是，它仍然使用append方法添加到列表中三次

newlist= [['id', 'name', 'lastContactedTime', 'email', 'phone_phones', 'home_phones', 'mobile_phones', 'work_phones', 'fax_phones', 'other_phones', 'address_1', 'address_2', 'address_3', 'city', 'state', 'postal_code', 'country', 'tags'], ['12-contacts', 'Courtney James', '', 'courtney@forlanchema.com', '+1 3455463849', '', '', '', '', '', '654 Rodney Franklin street', '', '', 'Birmingham', 'AL', '45678', 'US', ''], ['4-contacts', 'Joe Malcoun', '2019-08-13 14:41:12', 'ceo@nutshell.com', '', '', '', '', '', '', '212 South Fifth Ave', '', '', 'Ann Arbor', 'MI', '48103', 'US', ''], ['8-contacts', 'Rafael Acosta', '', 'racosta@forlanchema.com', '+1 338551534', '', '', '', '', '', '13 Jordan Avenue SW', '', '', 'Birmingham', 'AL', '45302', 'US', '']]
namelist = [] # new, empty list
for i in range(1, len(newlist)):
    names = newlist[i][1].split() # this yields [first_name, last_name]
    namelist.append([names[1], names[0]]) # [last_name, first_name]
companylist=[]
for i in range(1, len(newlist)):
    p = re.compile(r'(.+)@(.+)\.(.+)')
    test_str = newlist[i][3]
    company= re.findall(p, test_str)
    companyname= list(company[0][1])
    companynom=''.join(companyname)
    companylist.append(companynom) #yields company names
    # strip non-numeric characters'
    workphone = []
    wrkstreetaddress = []
    workcityaddress = []
    wrkstate = []
    wrkzip = []

for i in range(1, len(newlist)):
     phone = re.sub(r'\D', '', newlist[i][4])
 # remove leading 1 (area codes never start with 1)
     phone = phone.lstrip('1')
     workingphone= '{}.{}.{}'.format(phone[0:3], phone[3:6], phone[6:])
     workphone.append(workingphone) #yields a list of workphone numbers
     wrkstraddress= newlist[i][10]
     wrkstreetaddress.append(wrkstraddress) #yields a list of work street addresses
     wrkcityaddress= newlist[i][13] #yields a list of city addresses
     workcityaddress.append(wrkcityaddress)
     workstate= newlist[i][14] #yields a list of states
     wrkstate.append(workstate)
     workzip=newlist[i][15]
     wrkzip.append(workzip) #yields a list of zip codes

我希望每个列表都包含一个包含三项的列表：

如果我打印workstreetaddress列表，我会得到：

print(wrskstreetaddress)

['654 Rodney Franklin street', '212 South Fifth Ave', '13 Jordan Avenue SW']
instead of:
['654 Rodney Franklin street']
['654 Rodney Franklin street', '212 South Fifth Ave']
['654 Rodney Franklin street', '212 South Fifth Ave', '13 Jordan Avenue SW']

对于从companylist到wrkzip的所有其他列表也是一样的，我将这些项添加三次而不是一次，得到相同的结果

Answer 1

此print语句在代码末尾的结果：

print(workphone, wrkstreetaddress, workcityaddress, wrkstate, wrkzip)

收益率：

['345.546.3849', '..', '338.551.534'] ['654 Rodney Franklin street', '212 South Fifth Ave', '13 Jordan Avenue SW'] ['Birmingham', 'Ann Arbor', 'Birmingham'] ['AL', 'MI', 'AL'] ['45678', '48103', '45302']

我看你的单子没问题。你知道吗

Answer 2

有了pandas一切都会更好：

import pandas as pd

df = pd.DataFrame(newlist[1:], columns=newlist[0])

无需for-loops：

addresses = df.address_1.tolist()

print(addresses)

['654 Rodney Franklin street', '212 South Fifth Ave', '13 Jordan Avenue SW']

轻松添加或重命名df列：

# split name into first and last name
df[['first_name', 'last_name']] = df.name.str.split(' ', expand=True)

# rename id
df.rename(columns={'id': 'id'}, inplace=True)

# split country_code from phone_phones
df[['country_code', 'phone_phones']] = df.phone_phones.str.split(' ', expand=True)

现在数据将更易于处理。