Question

我正在尝试以以下格式打印字典。我想这样做是因为我需要使用JSON格式绘制D3中的一些趋势。对于这种趋势，我计算了每个州（1980年代至2010年代）在每个州的谋杀案数量。

我能够输出文件和所有内容，但是由于我试图创建图形，因此在标记文件中字典中的每个键，值对方面，JSON文件中的数据格式必须非常具体。输出。

xl = pd.ExcelFile('Wyoming.xlsx')
df = xl.parse('Sheet1')
year = df['Year']
state = df['State']
freq = dict()

for i in range(0, len(df)):
    currYear = year.iloc[i]
    if(currYear >= 1980 and currYear < 1989):
        currDecade = 1980
    elif(currYear >= 1990 and currYear < 2000):
        currDecade = 1990
    elif(currYear >= 2000 and currYear < 2010):
        currDecade = 2000
    elif(currYear >= 2010):
        currDecade = 2010
    currState = state.iloc[i]
    if currDecade in freq:
        if currState in freq[currDecade]:
            freq[currDecade][currState] += 1
        else:
            key = {currState: 1}
            freq[currDecade].update(key)
    else:
        key = {currDecade:{currState: 1}}
        freq.update(key)

#print(freq)
freq1 = [{'Decade': d, 'State': [{'State': s, 'Freq': f}]} for d, s, f in freq.items()]
print(freq1)

我收到错误消息“ ValueError：没有足够的值要解压（预期3，得到2）”

我希望输出如下。

[{"Decade": "1980", "State": [{"State": "California", "Freq": 29591}, {"State": "Massachusetts", "Freq": 1742}, ...}]

Answer 1

罪魁祸首是for d, s, f in freq.items()，因为freq.items()以频率返回(key, value)对中的可迭代对象。由于您已嵌套字典，请尝试以下操作：

freq1 = [{'Decade': d, 'State': [{'State': s, 'Freq': f} for s, f in sdict.items()]}
  for d, sdict in freq.items()
]

Answer 2

dict.items()仅使用两个元素来迭代元组：键和值。

freq1 = []
for decade, states in freq.items():
    freq1.append({
        'Decade': decade,
        'State': []
    })
    for state, freq in states.items():
        freq1['State'].append([{'State': state, 'Freq': freq}])
print(freq1)

我认为这种方式使代码更具可读性。但是，如果您仍然喜欢单行列表理解解决方案，则为：

freq1 = [{'Decade': d, 'State': [{'State': s, 'Freq': f} for s, f in ss.items()]} for d, ss in freq.items()]

ValueError：没有足够的值可解包（预期3，得到2）Python

2 个答案: