python去重（汉字一样，里面的编码不一样）？ python如何实现列表嵌套字典，字典内相同key去重，字典...
python\u4e2d\u5b9e\u73b0\u4e24\u4e2atxt\u6587\u6863\u7684\u53bb\u91cd\u529f\u80fd\uff08\u5373\u53bb\u6389\u7b2c\u4e00\u4e2a\u6587\u6863\u4e2d\u548c\u7b2c\u4e8c\u6587\u6863\u76f8\u540c\u7684\u884c\uff09# -*- coding: utf-8 -*-
#
# python 2.7
fp1 = file('', 'r')
fp2 = file('', 'r')
fp3 = file('', 'w')
d1 = {}
d2 = {}
isFirst = True
for line in fp1:
if not isFirst:
d1[hash(line)] = line
else:
isFirst = False
fp1.close()
isFirst = True
for line in fp2:
if not isFirst:
d2[hash(line)] = line
else:
isFirst = False
fp2.close()
diff = set(d1.keys()) - set(d2.keys())
for key in diff:
fp3.write(d1[key])
fp3.close()
\u6309\u7167\u4f60\u7684\u8981\u6c42\u7f16\u5199\u7684\u5b57\u5178\u5185\u76f8\u540ckey\u5408\u5e76\u7684Python\u7a0b\u5e8f\u5982\u4e0b
l=[{'a':1,'b':'haha'},{'a':3,'b':'haha'},{'a':2,'b':'xiaoming'}]
result=[]
temp=[]
for i in range(len(l)):
flag=False
suma=l[i]['a']
b=l[i]['b']
for j in range(i+1,len(l)):
if l[i]['b']==l[j]['b'] and (j not in temp):
flag=True
temp.append(i)
temp.append(j)
suma=suma+l[j]['a']
if i not in temp or flag==True:
result.append({'a':suma,'b':b})
print(result)
\u6e90\u4ee3\u7801(\u6ce8\u610f\u6e90\u4ee3\u7801\u7684\u7f29\u8fdb)
这是一个字符串的全角和半角的问题，可以导入 unicodedata 中的 normalize 函数先把全角转换为半角，然后再用 set 对列表去重，参考代码如下：
from unicodedata import normalize
list3=['热菜','凉菜','凉菜','硬菜']
set(map(lambda s: normalize('NFKC', s), list3))
输出：
{'硬菜', '热菜', '凉菜'}
扩展：
“ NFKC”代表“Normalization Form KC [Compatibility Decomposition, followed by Canonical Composition]”，并将全角字符替换为半角字符，这些半角字符与Unicode等价。
去重操作可以走个捷径，将列表转换为不重复的集合即可。示例代码见截图：
你说的‘凉菜’中的‘凉’字不一样，没看懂，字如果不一样在任何编码规则下其编码必须不同，这是基本的编码规则，不然电脑无法识别你要打出什么字了。
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