我有一个下面的表格列表:
<script type="text/javascript">
$("#submit").on("click",function() {
if ($("#title").val() != '') {
$("#loading").css("display", "none");
}
else {
$("#loading").css("display", "block");
setTimeout(function(){ $('#loading').fadeOut() }, 5000);
}
});
</script>
转换为
oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]
我试过了
final = [{'x':{'a':1,'b':2},'y':[2,3,2],'count':3},{'x':{'a':6,'b':7},'y':[2],'count':1},{'x':{'a':10,'b':11},'y':[4],'count':1}]
获取
s =设定([d [&#39; x&#39;]为旧名单中的d])
TypeError:不可用类型:&#39; dict&#39;
有更简单的方法吗?另外我知道x只能有三个值,所以我创建了三个变量list1,list2和list3。如果x可以有其他几个值,我必须找到类似的词典列表,如final!它也适用于字符串!
编辑:我试过这个。但这一切都搞砸了oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]
list1=[]
list2=[]
list3=[]
s = set([d['x'] for d in oldlist])
news=list(s)
for item in oldlist:
if item['x'] == news[0]:
list1.append(item['y'])
if item['x'] == news[1]:
list2.append(item['y'])
if item['x'] == news[2]:
list3.append(item['y'])
final=[]
dic1 = {'x':news[0],'y':list1,'count':len(list1)}
dic2 = {'x':news[1],'y':list2,'count':len(list2)}
dic3 = {'x':news[2],'y':list3,'count':len(list3)}
final.append(dic1)
final.append(dic2)
final.append(dic3)
print final
答案 0 :(得分:3)
set函数只能处理hashable
个对象,如字符串,数字,元组e.t.c
List,dict等数据类型是不可用的类型,因此set函数无法处理它们。
更清晰一点:
What do you mean by hashable in Python?
http://blog.lerner.co.il/is-it-hashable-fun-and-games-with-hashing-in-python/
您需要的基本实现:
for elem in oldlist:
found = False
for item in newlist:
if elem['x'] == item['x']:
y = item.get('y',[])
item['y'] = t.append(elem['y'])
found = True
break
if not found:
newlist.append({'x':elem['x'], 'y':[elem['y']]})
这将为您提供预期的结果
答案 1 :(得分:1)
设置python的功能不允许使用词典,你不能强制它,尝试另一种方法。 (仔细看看第5 和第6 行的评论)
试试这段代码:
oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]
list1=[]
list2=[]
list3=[]
s = [d['x'] for d in oldlist] # Placed the dictionaries in a list
s = result = [dict(tupleized) for tupleized in set(tuple(item.items()) for item in s)] # This is the manual way on removing duplicates dictionaries in a list instead of using set
news=list(s)
for item in oldlist:
if item['x'] == news[0]:
list1.append(item['y'])
if item['x'] == news[1]:
list2.append(item['y'])
if item['x'] == news[2]:
list3.append(item['y'])
final=[]
dic1 = {'x':news[0],'y':list1,'count':len(list1)}
dic2 = {'x':news[1],'y':list2,'count':len(list2)}
dic3 = {'x':news[2],'y':list3,'count':len(list3)}
final.append(dic1)
final.append(dic2)
final.append(dic3)
print final
答案 2 :(得分:1)
您可以使用defaultdict
其中键是frozenset
个对象,这些对象是从原始字母中的x
值创建的,值是相对y
的列表。然后,您可以使用列表推导构建最终结果,并将frozensets
转回dicts:
from collections import defaultdict
oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]
res = defaultdict(list)
for d in oldlist:
res[frozenset(d['x'].items())].append(d['y'])
final = [{'x': dict(k), 'y': v, 'count': len(v)} for k, v in res.items()] # [{'y': [2, 3, 2], 'x': {'a': 1, 'b': 2}, 'count': 3}, {'y': [4], 'x': {'a': 10, 'b': 11}, 'count': 1}, {'y': [2], 'x': {'a': 6, 'b': 7}, 'count': 1}]