将列表与 JSON 进行比较并计算出现次数答案

【问题标题】：Comparing list with JSON and counting occurences将列表与 JSON 进行比较并计算出现次数
【发布时间】：2021-08-10 13:13:02
【问题描述】：

给定列表如下：

make = ['ford', 'fiat', 'nissan', 'suzuki', 'dacia']
model = ['x', 'y', 'z']
version = ['A', 'B', 'C']
typ = ['sedan', 'coupe', 'van', 'kombi']
infos = ['steering wheel problems', 'gearbox problems', 'broken engine', 'throttle problems', None]

total.append(make)
total.append(model)
total.append(version)
total.append(typ)
total.append(infos)

我需要创建这些列表的所有可能组合的列表列表，所以我这样做了：

combos = list(itertools.product(*total))
all_combos = [list(elem) for elem in combos]

现在我想比较一下，在 JSON 对象中找到与 all_combos 的项目中出现的值相同的项目，并计算这些出现的次数。我的 JSON 很大，看起来有点像：

data = [
{  'make': 'dacia'
   'model': 'x',
   'version': 'A',
   'typ': 'sedan',
   'infos': 'steering wheel problems'
}, ...]

我想得到如下输出：

output = [
    {  'make': 'dacia'
       'model': 'x',
       'version': 'A',
       'typ': 'sedan',
       'infos': 'steering wheel problems',
       'number_of_occurences_of_such_combination_of_fields_with__such_values': 75
    }, ...]

如何解决这样的任务？

【问题讨论】：

标签： python json dictionary nested itertools

【解决方案1】：

如果我对您的理解正确，您希望在数据键 number_of_occurences_of_such_combination_of_fields_with__such_values 中添加每个字典：

from operator import itemgetter
from itertools import product

make = ["ford", "fiat", "nissan", "suzuki", "dacia"]
model = ["x", "y", "z"]
version = ["A", "B", "C"]
typ = ["sedan", "coupe", "van", "kombi"]
infos = [
    "steering wheel problems",
    "gearbox problems",
    "broken engine",
    "throttle problems",
    None,
]

total = [make, model, version, typ, infos]

data = [
    {
        "make": "dacia",
        "model": "x",
        "version": "A",
        "typ": "sedan",
        "infos": "steering wheel problems",
    },
    {
        "make": "dacia",
        "model": "x",
        "version": "A",
        "typ": "sedan",
        "infos": "steering wheel problems",
    },
    {
        "make": "ford",
        "model": "x",
        "version": "A",
        "typ": "sedan",
        "infos": "steering wheel problems",
    },
]

i = itemgetter("make", "model", "version", "typ", "infos")

cnt = {}
for c in itertools.product(*total):
    for d in data:
        if i(d) == c:
            cnt.setdefault(c, []).append(d)

for k, v in cnt.items():
    for d in v:
        d[
            "number_of_occurences_of_such_combination_of_fields_with__such_values"
        ] = len(v)

print(data)

打印：

[
    {
        "make": "dacia",
        "model": "x",
        "version": "A",
        "typ": "sedan",
        "infos": "steering wheel problems",
        "number_of_occurences_of_such_combination_of_fields_with__such_values": 2,
    },
    {
        "make": "dacia",
        "model": "x",
        "version": "A",
        "typ": "sedan",
        "infos": "steering wheel problems",
        "number_of_occurences_of_such_combination_of_fields_with__such_values": 2,
    },
    {
        "make": "ford",
        "model": "x",
        "version": "A",
        "typ": "sedan",
        "infos": "steering wheel problems",
        "number_of_occurences_of_such_combination_of_fields_with__such_values": 1,
    },
]

版本 2：（没有 itertools.product）：

from operator import itemgetter


i = itemgetter("make", "model", "version", "typ", "infos")

cnt = {}
for d in data:
    c = i(d)
    cnt[c] = cnt.get(c, 0) + 1

for d in data:
    d[
        "number_of_occurences_of_such_combination_of_fields_with__such_values"
    ] = cnt[i(d)]

print(data)

【讨论】：