Question

我是编程新手。我试图弄清楚如何从“实际”中减去“预算”，然后使用嵌套的for循环将值更新为“方差”。但是，我读到，迭代时更改字典不是最佳实践。到目前为止，我对如何继续感到困惑。

for i in properties:
    for j in properties[i]:
        if j == "actual":
            sum = properties[i][j]
            print('\nActual:' , sum)
        if j == "budgeted":
            sum_two = properties[i][j]
            print('Budgeted:' , sum_two)
            diff = sum_two - sum
            print('Variance:', diff)

default_value = 0

properties = {587: {'prop_name': 'Collington'}, 'rental_income': {'apartment_rent': '5120-0000', 'resident_assistance': '5121-0000', 'gain_loss': '5120-0000'}, 51200000: {'actual': 29620, 'budgeted': 30509, 'variance': default_value}, 51210000: {'actual': 25620, 'budgeted': 40509, 'variance': default_value}, ............

Answer 1

仅遍历字典并检查内部字典中是否存在actual，variance和budgeted，如果存在，则修改variance值< / p>

for k, v in properties.items():
    if (('actual' in v.keys()) and ('variance' in v.keys()) and ('budgeted' in v.keys())):
            properties[k]['variance'] = properties[k]['actual']-properties[k]['budgeted']

Answer 2

迭代时在字典中修改值并没有错。唯一不建议的方法是修改字典本身，即添加/删除元素。

Answer 3

尝试类似的东西：

for i in properties:
    properties[i]['variance'] = properties[i]['budgeted'] - properties[i]['actual']

如果不确定字典中是否存在bugeted和actual，则应捕获KeyError并进行适当处理：

for i in properties:
    try:
        properties[i]['variance'] = properties[i]['budgeted'] - properties[i]['actual']
    except KeyError:
        properties[i]['variance'] = -1 # Set to some special value or just pass

Answer 4

您的数据格式很奇怪，我总是尝试在字典中将类似的对象归为一组，而不是在字典的同一级别上包含元数据和项的“列表”。不过，这将为您工作：

for prop in properties:
    p = properties[prop]
    if 'actual' or 'budgeted' in p.keys():
        # get() wont error if not found, also default to 0 if not found
        p['variance'] = p.get('budgeted', 0) - p.get('actual', 0)

import json
print(json.dumps(properties, indent=4))

输出：

{
    "587": {
        "prop_name": "Collington"
    },
    "rental_income": {
        "apartment_rent": "5120-0000",
        "resident_assistance": "5121-0000",
        "gain_loss": "5120-0000"
    },
    "51200000": {
        "actual": 29620,
        "budgeted": 30509,
        "variance": 889
    },
    "51210000": {
        "actual": 25620,
        "budgeted": 40509,
        "variance": 14889
    }
}

Answer 5

sum = None
sum_two = None
for i in properties:
        for j in i:
            if j=="actual":
                sum = properties [i]["actual"] 
                print('\nActual:' , sum)
            if j == "budgeted":
                sum_two = properties[i]["budgeted"]
                print('Budgeted:' , sum_two)
                diff = sum_two - sum
                print('Variance:', diff)

我没有得到确切的含义，但是应该可以。

循环嵌套字典

5 个答案: