python添加两个列表并创建一个新的lis

Question

lst1 = ['company1,AAA,7381.0 ', 'company1,BBB,-8333.0 ', 'company1,CCC,
3079.999 ', 'company1,DDD,5699.0 ', 'company1,EEE,1640.0 ',
       'company1,FFF,-600.0 ', 'company1,GGG,3822.0 ', 'company1,HHH,-600.0 ',
       'company1,JJJ,-4631.0 ', 'company1,KKK,-400.0 ']

lst2 =['company1,AAA,-4805.0 ', 'company1,ZZZ,-2576.0 ', 'company1,BBB,1674.0 ', 'company1,CCC,3600.0 ', 'company1,DDD,1743.998 ']

我需要的输出==

['company1,AAA,2576.0','company1,ZZZ,-2576.0 ','company1,KKK,-400.0 ' etc etc]

我需要在每个列表中添加相似的产品编号，并将其移动到新列表中。我还需要任何符号没有被添加到一起被添加到新的名单。我在浏览每个列表时遇到问题。你知道吗

这就是我所拥有的：

h = [] 

z = []         

a = []        

for g in lst1:
    spl1 = g.split(",")
    h.append(spl1[1])
for j in lst2:
    spl2 = j.split(",")
    **if spl2[1] in h:
        converted_num =(float(spl2[2]) +float(spl1[2]))
        pos=('{0},{1},{2}'.format(spl2[0],spl2[1],converted_num))
        z.append(pos)**
    else:
        pos=('{0},{1},{2}'.format(spl2[0],spl2[1],spl2[2]))
        z.append(pos)

for f in z:
    spl3 = f.split(",")
    a.append(spl3[1])

for n in lst1[:]:
    spl4 = n.split(",")
    if spl4[1] in a:
        got = (spl4[0],spl4[1],spl4[2])
        lst1.remove(n)
smash = lst1+z #for i in smash:
for i in smash:
    print(i)

我在遍历列表以确保将所有simliar产品放到一个新列表（粗体）时遇到问题，并且将不在列表1中但在lst2中的任何产品放到新列表中，反之亦然。我相信有一个更简单的方法。你知道吗

Answer 1

完全解决你的问题

from collections import OrderedDict
lst1 = ['company1,AAA,7381.0 ', 'company1,BBB,-8333.0 ', 'company1,CCC, 3079.999 ', 
        'company1,DDD,5699.0 ', 'company1,EEE,1640.0 ', 'company1,FFF,-600.0 ', 
        'company1,GGG,3822.0 ', 'company1,HHH,-600.0 ', 'company1,JJJ,-4631.0 ', 
        'company1,KKK,-400.0 ']
lst2 =['company1,AAA,-4805.0 ', 'company1,ZZZ,-2576.0 ', 'company1,BBB,1674.0 ', 
       'company1,CCC,3600.0 ', 'company1,DDD,1743.998 ']
#Create an OrderedDict, this will ensure that the Order in the original
# List is maintained
out_dict = OrderedDict()
#Create a Dictionary out of the first List, Note the key is
#the company and product ID
for e in lst1:
    key, _, value = e.strip().rpartition(",")
    out_dict[key] = float(value)
#Now iterate through the next list
for e in lst2:
# Partition on company, productID and Value
    key, _, value = e.strip().rpartition(",")
#Retrieve the value with the key with default '0'
#and add value from the new list
    out_dict[key] = out_dict.get(key, 0) + float(value)
#Finally recreate the list from the dictionary
lst3 = [','.join(map(str, e)) for e in out_dict.iteritems()]
print lst3

Answer 2

我赞成使用dict的建议。因为我很懒，所以我更喜欢使用defaultdict，因为这样我就不用担心检查键是否存在了。（您也可以在这里使用Counter。）特别是，假设您必须以列表开始和结束：

from collections import defaultdict

data = defaultdict(float)

for line in lst1+lst2:
    name, code, value = line.split(",")
    data[name, code] += float(value)

newlist = ['{},{},{}'.format(key[0], key[1], val) for key, val in sorted(data.items())]

给予

>>> data
defaultdict(<type 'float'>, {('company1', 'HHH'): -600.0, ('company1', 'JJJ'): -4631.0,
('company1', 'KKK'): -400.0, ('company1', 'DDD'): 7442.998, ('company1', 'ZZZ'): -2576.0,
('company1', 'CCC'): 6679.999, ('company1', 'AAA'): 2576.0, ('company1', 'FFF'): -600.0,
('company1', 'GGG'): 3822.0, ('company1', 'EEE'): 1640.0, ('company1', 'BBB'): -6659.0})

以及

>>> newlist
['company1,AAA,2576.0', 'company1,BBB,-6659.0', 'company1,CCC,6679.999',
'company1,DDD,7442.998', 'company1,EEE,1640.0', 'company1,FFF,-600.0', 
'company1,GGG,3822.0', 'company1,HHH,-600.0', 'company1,JJJ,-4631.0', 
'company1,KKK,-400.0', 'company1,ZZZ,-2576.0']

Answer 3

您应该使用字典将字符串映射到浮点数，如下所示：

dict1 = {'company1,AAA':7381.0, 'company1,BBB':-8333.0, 'company1,CCC':3079.999}

当你得到一个新的口述：

dict2 = {'company1,AAA':-4805.0}

你可以说：

for key in dict2.keys():
    dict1[key] = dict1[key] + dict2[key]

或者类似的东西（另请参见function.update（））

编辑：

update（）函数还允许您使用dict2中的新项更新字典，只需使用一个简单的if语句检查dict2中的键是否还不在dict1中