嵌套字典蟒蛇键
我想是以下结构:y['1'][tuple(list)] = val
作为一个Python嵌套字典,但同时,我努力,我得到KeyError
:在csv文件嵌套字典蟒蛇键
的数据是这样的:
Rest_id, rates, items 1,4, burger
1,8, tofu_log
2,5, burger
2,8.5, tofu_log
3,4, chef_salad
3,8, steak_salad_sandwich
4,5, steak_salad_sandwich,salad
4,2.5, wine_spritzer
5,4, extreme_fajita3,test2,test4,test
5,8, fancy_european_water
6,5, fancy_european_water
6,6, extreme_fajita, jalapeno_poppers, extra_salsa
7,1.5, wine_spritzer
7,5, extreme_fajita, jalapeno_poppers
以下是代码:
y = defaultdict(dict) with open('sample_data_tested_with.csv','r') as f:
reader = csv.reader(f, delimiter=',', quoting=csv.QUOTE_NONE)
reader = [[x.strip() for x in row] for row in reader]
for i in reader:
#cd[i[0]] = {tuple(i[2:]):i[1]}
#cd[i[0]][tuple(i[2:])].update(i[1])
print i[0], i[1], tuple(i[2:])
y[i[0]][tuple(i[2:])].append(i[1])
后来我想在这样的y['rest_id']['item']
字典搜索并找到该利率。 在此先感谢。
从IPython的完整的堆栈:
In [49]: with open('sample_data_tested_with.csv','r') as f: reader = csv.reader(f, delimiter=',', quoting=csv.QUOTE_NONE)
reader = [[x.strip() for x in row] for row in reader]
for i in reader:
#cd[i[0]] = {tuple(i[2:]):i[1]}
#cd[i[0]][tuple(i[2:])].update(i[1])
print i[0], i[1], tuple(i[2:])
#x[tuple(i[2:])]=float(i[1])
y[i[0]][tuple(i[2:])].append(i[1])
....:
1 4 ('burger',)
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-49-ab608c2dc33a> in <module>()
7 print i[0], i[1], tuple(i[2:])
8 #x[tuple(i[2:])]=float(i[1])
----> 9 y[i[0]][tuple(i[2:])].append(i[1])
10
KeyError: ('burger',)
回答:
看起来你可以使用itertools.groupby
>>>import itertools >>>newreader,keys=[],[]
>>>for k,g in itertools.groupby(reader,key=lambda x:x[0]):
newreader.append(tuple(g[1:]))
keys.append(k)
这应该您的数据组到元组。 现在我们遍历newreader将每个元组转换为字典。
>>>d={} >>>for *i,j in newreader,keys:
d[j]=dict(i)
回答:
下面是我得到的解决方案,请检查并让我们知道是否有人发现任何问题,非常感谢。
def create_NestedDict(filename): NestedDict = defaultdict(dict)
with open(filename,'rb') as f:
reader = csv.reader(f, delimiter=',')
for row in reader:
#print "row :", row
record = row
if record[0] not in NestedDict:
NestedDict[record[0]] = {}
items = tuple([ i.strip() for i in record[2:]])
[record[0]][items] = record[1]
return NestedDict
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