Question

简介：

我正在尝试访问python中的嵌套字典元素，如下所示：

{'CA':{'1':'3','2':'3','3':'3'},'IL': {'1':'31','2':'45','3':'23'},...}

首先，我从Excel文件中读取，我获取状态名称，然后为每个州分配字典。我就是这样做的：

xls_file = pd.ExcelFile('D:/CollegeScorecardDataDictionary-08-18-2016.xlsx')
dfEx = xls_file.parse('Church')  # Parse Church sheet
for (i, item) in enumerate(stateName):
     if stateChurchDict.get(item)<>None:
            continue 
     else:
            stateChurchDict[item] = dict

一旦迭代循环，我就会这样：

{'CA':<type dict>,'IL': <type dict>,'AL': <type dict>...}

在每个州都有很多教堂可以归类为'1', '2' or '3' 这是我在嵌套字典中获得数字的地方。

我的问题是我想引用某些状态的嵌套字典，如

stateChurchDict['AL']['3']

并获得类别＆＃39; 3＆＃39;下的教堂数量在某种状态下。但是，首先我必须检查它是否为空，如果它是空的，则必须添加该值。因此，我想出了这个：

for (i, item) in enumerate(stateName):
            if stateChurchDict[stateName[i-1]]['3'] <> None:
                stateChurchDict.update({stateChurchDict[stateName[i-1]]['3']: stateChurchDict[stateName[i-1]]['3'] + 1})
            else:
                stateChurchDict[stateName[i-1]]['3'] = 1

但是，stateChurchDict[stateName[i-1]]['3']这个stateName[i-1] == 'AL'无法访问嵌套字典stateChurchDict['AL']['3']，它会调用像import pandas as pd from collections import defaultdict, Counter def IsNumeric(x): try: float(x) return x except: return 0 xls_file = pd.ExcelFile('D:/CollegeScorecardDataDictionary-08-18-2016.xlsx') dfEx = xls_file.parse('Church') # Parse data_dictionary sheet dfCsv = pd.read_csv('D:/MERGED2014_15_PP.csv', low_memory=False) churchCode = dfEx.Code # Label column churchName = dfEx.ChurchName # Value column churchCategory = dfEx.Category # Church category relafil = dfCsv.RELAFFIL # Religious Id in CSV stateName = dfCsv.STABBR # Name of state churchList = {} # Create dictionary to store churches stateChurchDict = defaultdict(Counter) # Create dictionary to store churches by state stateChurchTemp = {} #Sepate dictionary for churches by state # Put values into dictionary for (i, v) in enumerate(churchCode): churchList[v] = churchCategory[i] #Assigns a category to each church in state for (i, item) in enumerate(stateName): #Create a dictionary as a value to each state in the stateChurchList dictionary if item <> None: if stateChurchDict.get(item) <> None: continue else: stateChurchDict[item] = {} for (i, item) in enumerate(stateName): #Iterate through states and count the number of churches by categories. Once the state name is changed, the number needs to be transferred from stateChurchTemp to stateChurchDict if IsNumeric(relafil[i]) <> 0: if i >= 1 and item <> stateName[i - 1]: if stateChurchDict[stateName[i - 1]][3] <> None: stateChurchDict.update({stateChurchDict[stateName[i - 1]][3]: stateChurchDict[stateName[i - 1]][ 3] + IsNumeric( stateChurchTemp[3])}) else: stateChurchDict[stateName[i - 1]][3] = IsNumeric(stateChurchTemp[3]) if stateChurchDict[stateName[i - 1]][2] <> None: stateChurchDict.update({stateChurchDict[stateName[i - 1]][2]: stateChurchDict[stateName[i - 1]][ 2] + IsNumeric( stateChurchTemp[2])}) else: stateChurchDict[stateName[i - 1]][2] = IsNumeric(stateChurchTemp[2]) if stateChurchDict[stateName[i - 1]][1] <> None: stateChurchDict.update({stateChurchDict[stateName[i - 1]][1]: stateChurchDict[stateName[i - 1]][ 1] + IsNumeric( stateChurchTemp[1])}) else: stateChurchDict[stateName[i - 1]][1] = IsNumeric(stateChurchTemp[1]) if churchList.get(relafil[i]) <> None and stateChurchTemp.get(churchList.get(relafil[i])) <> None: stateChurchTemp.update({churchList.get(relafil[i]): stateChurchTemp.get(churchList.get(relafil[i])) + 1}) else: stateChurchTemp[churchList.get(relafil[i])] = 1 print stateChurchDict这样的元素，但仍然没有。

非常感谢任何帮助。

我发布了所有内容以获得更好的解释：

def reverse(arr):
    for i in range(len(arr) / 2): 
        arr[-i-1], arr[i] = arr[i], arr[-i-1]

Answer 1

stateChurchDict首先应该不是 vanilla 字典。

传递给collections.Counter的collections.defauldict是正确的方向：

>>> from collections import defaultdict, Counter
>>> d = defaultdict(Counter) # or defaultdict(lambda: defaultdict(int))
>>> d['AL']['3']
0

使用此功能，可以动态生成默认计数值，以便最多嵌套两个级别的密钥，并且您的代码将减少为：

from collections import defaultdict, Counter

stateChurchDict = defaultdict(Counter)
for i, item in enumerate(stateName):
    stateChurchDict[stateName[i-1]]['3'] += 1

Answer 2

好的，这就是我的回答。简单地说，我创建了另一个字典internalDict，它存储了嵌套字典的值。

我没有致电stateChurchDict[stateName[i-1][3]]，而是尝试了internalDict = stateChurchDict[stateName[i-1]]，之后只与internalDict合作。因此，调用stateChurchDict[stateName[i-1][3]]就像internalDict[3]。

Answer 3

您没有调用嵌套字典，而是尝试更新主字典。请更改此行：

stateChurchDict.update({stateChurchDict[stateName[i - 1]][2]: stateChurchDict[stateName[i - 1]][2] + IsNumeric(stateChurchTemp[2])})

用这个：

stateChurchDict.get(statename[i-1]).update({3: stateChurchDict[stateName[i - 1]][3] + IsNumeric(stateChurchTemp[3])})

无法在python中访问嵌套字典

3 个答案: