函數 criarB 返回：{269476: 200129, 209624: 200129 ...} 200129 position in aux2, '200129 and 209624 are keys'如何根據鍵拆分dict，如果keys % 10 = 0 必須存儲在list[0]中list = [[], [], ...]類完成：class Buckets:def __init__(self, keys, palavras, tamanhoP): self.listaBuckts = dict() self.listaHash = list() self.keys = list(keys) aux = list(zip(keys, palavras)) self.aux2 = list() for i in range(0, len(aux), tamanhoP): self.aux2.append(dict(aux[i:i + tamanhoP]))def criarB(self): for i, pag in enumerate(self.aux2): for v in pag.keys(): self.listaBuckts[v] = i return self.listaBucktsdef indexar(self): count = 0 buckets = [[] for _ in range(10)] for r in range(0, len(buckets)): for s in range(0, len(self.listaBuckts)): if s % 3 == 0: buckets[r].append([v + count for v in self.listaHash[s:s + 3]]) count += 1 return buckets[0]“索引”功能使用一個只包含索引的列表，如何使用 self.buckets 字典的鍵并根據功能劃分字典試圖：class Buckets:def __init__(self, keys, palavras, tamanhoP): self.listaBuckts = dict() self.listaHash = list() self.keys = list(keys) aux = list(zip(keys, palavras)) self.aux2 = list() for i in range(0, len(aux), tamanhoP): self.aux2.append(dict(aux[i:i + tamanhoP]))def criarB(self): for i, pag in enumerate(self.aux2): for v in pag.keys(): self.listaBuckts[v] = i return self.listaBucktsdef indexar(self): test = [[] for _ in range(10)] for x in self.listaBuckts: i = x % 10 test[i].append([x, dict[x]]) return test[0]
2 回答
GCT1015
TA貢獻1827條經驗 獲得超4個贊
您可以使用列表理解：
listOfDicts = [{k:v for k,v in dictionary.items() if k%10==i} for i in range(10)]
或循環：
listOfDicts = [ dict() for _ in range(10) ]
for key,value in dictionary.items():
listOfDicts[key%10].update({key:value})
繁華開滿天機
TA貢獻1816條經驗 獲得超4個贊
如果要根據鍵將字典拆分為兩個列表。我認為以下可以工作
dict #The dictionary returned from function crairB
a = [] #List to save keys and values for key%10==0
b = [] #List for key%10!=0
for x in dict:
if x%10==0:
a.append([x,dict[x]]) #Here I have saved both keys and values in the list, you can edit according to your need.
else:
b.append([x,dict[x]])
編輯：
如果您必須將數據拆分為行等于鍵 % 10 的二維列表，那么我認為以下方法可以工作。
list = [[] for _ in range(10)]
for x in dict:
i = x%10
list[i].append([x,dict[x]])
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