我有一个列表列表，看起来像这样，是从格式不正确的csv文件中提取的：
DF = [['Customer Number: 001 '],
['Notes: Bought a ton of stuff and was easy to deal with'],
['Customer Number: 666 '],
['Notes: acted and looked like Chris Farley on that hidden decaf skit from SNL'],
['Customer Number: 103 '],
['Notes: bought a ton of stuff got a free keychain'],
['Notes: gave us a referral to his uncles cousins hairdresser'],
['Notes: name address birthday social security number on file'],
['Customer Number: 007 '],
['Notes: looked a lot like James Bond'],
['Notes: came in with a martini']]
我想得到这样的新结构：
['Customer Number: 001 Notes: Bought a ton of stuff and was easy to deal with',
'Customer Number: 666 Notes: acted and looked like Chris Farley on that hidden decaf skit from SNL',
'Customer Number: 103 Notes: bought a ton of stuff got a free keychain',
'Customer Number: 103 Notes: gave us a referral to his uncles cousins hairdresser',
'Customer Number: 103 Notes: name address birthday social security number on file',
'Customer Number: 007 Notes: looked a lot like James Bond',
'Customer Number: 007 Notes: came in with a martini']
之后我可以进一步分裂、剥离等。
因此，我使用了以下事实：
- 客户编号始终以
Customer Number开头 Notes总是更长Notes数量永远不会超过5
编写显然是荒谬的解决方案，即使它有效。
DF = [item for sublist in DF for item in sublist]
DF = DF + ['stophere']
DF2 = []
for record in DF:
if (record[0:17]=="Customer Number: ") & (record !="stophere"):
DF2.append(record + DF[DF.index(record)+1])
if len(DF[DF.index(record)+2]) >21:
DF2.append(record + DF[DF.index(record)+2])
if len(DF[DF.index(record)+3]) >21:
DF2.append(record + DF[DF.index(record)+3])
if len(DF[DF.index(record)+4]) >21:
DF2.append(record + DF[DF.index(record)+4])
if len(DF[DF.index(record)+5]) >21:
DF2.append(record + DF[DF.index(record)+5])
有人介意为此类问题推荐更稳定、更智能的解决方案吗？
只需跟踪我们何时找到新客户：
from pprint import pprint as pp
out = []
for sub in DF:
if sub[0].startswith("Customer Number"):
cust = sub[0]
else:
out.append(cust + sub[0])
pp(out)
输出：
['Customer Number: 001 Notes: Bought a ton of stuff and was easy to deal with',
'Customer Number: 666 Notes: acted and looked like Chris Farley on that '
'hidden decaf skit from SNL',
'Customer Number: 103 Notes: bought a ton of stuff got a free keychain',
'Customer Number: 103 Notes: gave us a referral to his uncles cousins '
'hairdresser',
'Customer Number: 103 Notes: name address birthday social security number '
'on file',
'Customer Number: 007 Notes: looked a lot like James Bond',
'Customer Number: 007 Notes: came in with a martini']
如果客户稍后可以再次重复，并且您希望将它们组合在一起，请使用字典：
from collections import defaultdict
d = defaultdict(list)
for sub in DF:
if sub[0].startswith("Customer Number"):
cust = sub[0]
else:
d[cust].append(cust + sub[0])
print(d)
输出：
pp(d)
{'Customer Number: 001 ': ['Customer Number: 001 Notes: Bought a ton of '
'stuff and was easy to deal with'],
'Customer Number: 007 ': ['Customer Number: 007 Notes: looked a lot like '
'James Bond',
'Customer Number: 007 Notes: came in with a '
'martini'],
'Customer Number: 103 ': ['Customer Number: 103 Notes: bought a ton of '
'stuff got a free keychain',
'Customer Number: 103 Notes: gave us a referral '
'to his uncles cousins hairdresser',
'Customer Number: 103 Notes: name address '
'birthday social security number on file'],
'Customer Number: 666 ': ['Customer Number: 666 Notes: acted and looked '
'like Chris Farley on that hidden decaf skit '
'from SNL']}
根据您的评论和错误，您似乎在实际客户之前有行，因此我们可以将它们添加到列表中的第一个客户：
# added ["foo"] before we see any customer
DF = [["foo"],['Customer Number: 001 '],
['Notes: Bought a ton of stuff and was easy to deal with'],
['Customer Number: 666 '],
['Notes: acted and looked like Chris Farley on that hidden decaf skit from SNL'],
['Customer Number: 103 '],
['Notes: bought a ton of stuff got a free keychain'],
['Notes: gave us a referral to his uncles cousins hairdresser'],
['Notes: name address birthday social security number on file'],
['Customer Number: 007 '],
['Notes: looked a lot like James Bond'],
['Notes: came in with a martini']]
from pprint import pprint as pp
from itertools import takewhile, islice
# find lines up to first customer
start = list(takewhile(lambda x: "Customer Number:" not in x[0], DF))
out = []
ln = len(start)
# if we had data before we actually found a customer this will be True
if start:
# so set cust to first customer in list and start adding to out
cust = DF[ln][0]
for sub in start:
out.append(cust + sub[0])
# ln will either be 0 if start is empty else we start at first customer
for sub in islice(DF, ln, None):
if sub[0].startswith("Customer Number"):
cust = sub[0]
else:
out.append(cust + sub[0])
哪些输出：
['Customer Number: 001 foo',
'Customer Number: 001 Notes: Bought a ton of stuff and was easy to deal with',
'Customer Number: 666 Notes: acted and looked like Chris Farley on that '
'hidden decaf skit from SNL',
'Customer Number: 103 Notes: bought a ton of stuff got a free keychain',
'Customer Number: 103 Notes: gave us a referral to his uncles cousins '
'hairdresser',
'Customer Number: 103 Notes: name address birthday social security number '
'on file',
'Customer Number: 007 Notes: looked a lot like James Bond',
'Customer Number: 007 Notes: came in with a martini']
我假设您会认为任何客户之前的行实际上属于第一个客户。
您的基本目标是对注释进行分组并将其与客户相关联。而且由于列表已经排序，您可以简单地使用 itertools.groupby ，像这样
from itertools import groupby, chain
def build_notes(it):
customer, func = "", lambda x: x.startswith('Customer')
for item, grp in groupby(chain.from_iterable(DF), key=func):
if item:
customer = next(grp)
else:
for note in grp:
yield customer + note
# In Python 3.x, you can simply do
# yield from (customer + note for note in grp)
在这里，我们将实际的列表列表扁平化为字符串序列，并带有 chain.from_iterable .然后我们将有Customer的行和没有的行分组。如果该行有 Customer ，则item将被True否则False。如果item是True，那么我们得到客户信息，当item False时，我们迭代分组的笔记，并通过将客户信息与笔记连接起来一次返回一个字符串。
因此，当您运行代码时，
print(list(build_notes(DF)))
你得到
['Customer Number: 001 Notes: Bought a ton of stuff and was easy to deal with',
'Customer Number: 666 Notes: acted and looked like Chris Farley on that hidden decaf skit from SNL',
'Customer Number: 103 Notes: bought a ton of stuff got a free keychain',
'Customer Number: 103 Notes: gave us a referral to his uncles cousins hairdresser',
'Customer Number: 103 Notes: name address birthday social security number on file',
'Customer Number: 007 Notes: looked a lot like James Bond',
'Customer Number: 007 Notes: came in with a martini']
DF = [['Customer Number: 001 '],
['Notes: Bought a ton of stuff and was easy to deal with'],
['Customer Number: 666 '],
['Notes: acted and looked like Chris Farley on that hidden decaf skit from SNL'],
['Customer Number: 103 '],
['Notes: bought a ton of stuff got a free keychain'],
['Notes: gave us a referral to his uncles cousins hairdresser'],
['Notes: name address birthday social security number on file'],
['Customer Number: 007 '],
['Notes: looked a lot like James Bond'],
['Notes: came in with a martini']]
custnumstr = None
out = []
for df in DF:
if df[0].startswith('Customer Number'):
custnumstr = df[0]
else:
out.append(custnumstr + df[0])
for e in out:
print e
您还可以使用 OrderedDict，其中键是客户，值是注释列表：
from collections import OrderedDict
DF_dict = OrderedDict()
for subl in DF:
if 'Customer Number' in subl[0]:
DF_dict[subl[0]] = []
continue
last_key = list(DF_dict.keys())[-1]
DF_dict[last_key].append(subl[0])
for customer, notes in DF_dict.items():
for a_note in notes:
print(customer,a_note)
结果：
Customer Number: 001 Notes: Bought a ton of stuff and was easy to deal with
Customer Number: 666 Notes: acted and looked like Chris Farley on that hidden decaf skit from SNL
Customer Number: 103 Notes: bought a ton of stuff got a free keychain
Customer Number: 103 Notes: gave us a referral to his uncles cousins hairdresser
Customer Number: 103 Notes: name address birthday social security number on file
Customer Number: 007 Notes: looked a lot like James Bond
Customer Number: 007 Notes: came in with a martini
客户的注释数量、计算注释或仅为给定客户选择注释，则在这样的字典中输入值可能很有用。
或者，在每次迭代中无需调用list(DF_dict.keys())[-1]：
last_key = ''
for subl in DF:
if 'Customer Number' in subl[0]:
DF_dict[subl[0]] = []
last_key = subl[0]
continue
DF_dict[last_key].append(subl[0])
和新的较短版本，使用默认：
from collections import defaultdict
DF_dict = defaultdict(list)
for subl in DF:
if 'Customer Number' in subl[0]:
customer = subl[0]
continue
DF_dict[customer].append(subl[0])
只要格式与您的示例相同，这应该有效。
final_list = []
for outer_list in DF:
for s in outer_list:
if s.startswith("Customer"):
cust = s
elif s.startswith("Notes"):
final_list.append(cust + s)
for f in final_list:
print f
只要你能指望第一个元素是客户，你就可以这样做。
只需遍历每个项目。如果物料是客户，请将当前客户设置为该字符串。否则，它是一个注释，因此您将客户和注释附加到结果列表中。
customer = ""
results = []
for record in DF:
data = record[0]
if "Customer" in data:
customer = data
elif "Notes" in data:
result = customer + data
results.append(result)
print(results)