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 ']
输出我需要==
['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类似的产品编号并将其移至新列表。我还需要将任何符号添加到一起添加到新列表中。我在浏览每个列表时遇到问题。
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)
我在遍历列表时遇到问题,以确保我将所有simliar产品都添加到新列表(粗体)和任何不在列表1中但在lst2中的产品到新列表,反之亦然。我相信有一个更简单的方法。
答案 0 :(得分:2)
我是第二个使用-dict的建议。因为我很懒,所以我更喜欢使用defaultdict
,因为我不必担心检查密钥是否存在。 (您也可以在这里使用Counter
。)特别是,假设您必须以列表开头和结尾:
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())]
给出
>>> 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})
和
>>> 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']
答案 1 :(得分:1)
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}
然后当你得到一个新的词典时:
dict2 = {'company1,AAA':-4805.0}
你可以说:
for key in dict2.keys():
dict1[key] = dict1[key] + dict2[key]
或沿着这些方向的东西(也参见函数.update())
编辑:
.update()函数还允许您使用dict2中的新项更新字典,只使用简单的if语句检查dict2中的键是否已经存在于dict1中
答案 2 :(得分:0)
问题的完整解决方案
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