Question

我有一个列表，看起来像这样，是从格式不佳的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']]

我想最终得到一个像这样的新结构：

['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']

之后我可以进一步拆分，剥离等。

所以，我使用了以下事实：

客户编号始终以Customer Number
Notes总是更长
Notes的数量永远不会超过5

编写一个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])

有人会介意为这类问题推荐更稳定，更智能的解决方案吗？

Answer 1

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']

如果客户可以稍后重复，并且您希望将它们组合在一起，请使用dict：

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线路，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])

哪个输出：

 ['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']

我认为你会考虑在任何客户真正属于第一个客户之前的线路。

Answer 2

您的基本目标是将笔记分组并与客户关联。由于列表已经排序，您只需使用itertools.groupby，就像这样

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展平为一系列字符串，chain.from_iterable。然后我们将其中包含Customer的行和不包含Customer的行分组。如果该行有item，则True将为False，否则为item。如果True为item，那么我们会获取客户信息，当False为print(list(build_notes(DF)))时，我们会对分组的备注进行迭代，并通过连接一次返回一个字符串客户信息与笔记。

所以，当你运行代码时，

['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']

你得到了

{{1}}

Answer 3

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

Answer 4

您也可以使用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

如果您想计算给定客户的笔记数量，计算笔记，或者只选择给定客户的笔记，那么将值放在这样的字典中会非常有用。

替代方案，没有在每次迭代中调用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])

新的更短版本，使用defaultdict：

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])

Answer 5

只要格式与您的示例相同，这应该可以。

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

Answer 6

只要您可以指望第一个元素是客户，您就可以这样做。

只需遍历每个项目。如果该项目是客户，请将当前客户设置为该字符串。另外，这是一个注释，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)

遍历循环中的索引列表，以重新格式化字符串

6 个答案: