Python: 如果其他值为真，则计算值在嵌套字典中的出现次数

Question

我有一个嵌套字典，如下所示：

{
1: {'Name': {'John'}, 'Answer': {'yes'}, 'Country': {'USA'}}, 
2: {'Name': {'Julia'}, 'Answer': {'no'}, 'Country': {'Hong Kong'}}
3: {'Name': {'Adam'}, 'Answer': {'yes'}, 'Country': {'Hong Kong'}}
}

我现在需要得到每个国家出现的次数以及回答是或否的人数。目前我只收集每个国家出现的次数：

nationalities = ['USA', 'Hong Kong', 'France' ...]
for countries in nationalities:
    cnt =[item for l in [v2 for v1 in dictionary1.values() for v2 in v1.values()] for item in l].count(countries)
    result.append(countries + ': ' + str(cnt))

所以使用我的数据表我得到类似

的东西

['Hong Kong: 2', 'France: 2', 'Italy: 3']

但是，我想得到回答是和回答不是的人的比例。这样我得到一个 ['Hong Kong: 2 1 1'] 形式的列表，其中第一个数字是总数，第二个和第三个分别是 yes 和 no

感谢您的帮助

Answer 1

a={
1: {'Name': {'John'}, 'Answer': {'yes'}, 'Country': {'USA'}}, 
2: {'Name': {'Julia'}, 'Answer': {'no'}, 'Country': {'Hong Kong'}},
3: {'Name': {'Adam'}, 'Answer': {'yes'}, 'Country': {'Hong Kong'}}
}
results=[]
nationalities = ['USA', 'Hong Kong', 'France']
for country in nationalities:
    countryyes=0
    countryno=0
    for row in a.values():
        if str(row['Country'])[2:-2] == country:
            if str(row['Answer'])[2:-2] == 'yes':
                countryyes+=1
            if str(row['Answer'])[2:-2] == 'no':
                countryno+=1
    results.append(country+': '+str(countryyes+countryno)+' '+str(countryyes)+' '+str(countryno))

我想做一些笔记。首先，我将 countries 改为 country（这样在 for 循环中使用复数名称是不正常的）。其次，我想评论一下，如果您上面的代码在一组中有名称、答案和国家/地区，我认为您最好将其更改为仅将其作为字符串。

Answer 2

这是一个可能的解决方案，它使用 defaultdict 来生成结果字典，方法是对每个 country 求和多少等于 yes 或 no 的答案：

from collections import defaultdict

dictionary1 = {
1: {'Name': {'John'}, 'Answer': {'yes'}, 'Country': {'USA'}}, 
2: {'Name': {'Julia'}, 'Answer': {'no'}, 'Country': {'Hong Kong'}},
3: {'Name': {'Adam'}, 'Answer': {'yes'}, 'Country': {'Hong Kong'}}
}

nationalities = ['USA', 'Hong Kong', 'France']
result = defaultdict(list)
for countries in nationalities:
    [yes, no] = [sum(list(d['Answer'])[0] == answer and list(d['Country'])[0] == countries for d in dictionary1.values()) for answer in ['yes', 'no']]
    result[countries] = [ yes+no, yes, no ]
    
print(dict(result))

对于您的示例数据，这给出了

{
 'USA': [1, 1, 0],
 'Hong Kong': [2, 1, 1],
 'France': [0, 0, 0]
}

然后您可以通过

将其转换为字符串列表

result = [ f"{key}: {' '.join(map(str, counts))}" for key, counts in result.items()]

给出：

['USA: 1 1 0', 'Hong Kong: 2 1 1', 'France: 0 0 0']

Answer 3

我会使用 Counter to count answers and groupby() 按国家/地区对条目进行分组：

from collections import Counter
from operator import itemgetter
from itertools import groupby

dictionary1 = {...}  # input data
group_func = itemgetter('Country')
result = []
for (country, *_), items in groupby(sorted(dictionary1.values(), key=group_func), group_func):
    answers = Counter(answer.lower() for i in items for answer in i['Answer'])
    result.append(f'{country} {sum(answers.values())} {answers.get("yes", 0)} {answers.get("no", 0)}')

Python: 如果其他值为真，则计算值在嵌套字典中的出现次数

Python: Count the occurence of value in nested dictionary if other value is true

python

dictionary

for-loop

list

find-occurrences