格式化要在图表中显示的数据点列表

Format list of data points to display in graph

我正在 Django Rest Framework 中编写一个 API 端点,并希望为图形格式化数据。我从数据库中得到这样的数据:

data = [
    {
        "name": "Test 3",
        "status": "Active",
        "count": 1
    },
    {
        "name": "Test 2",
        "status": "Failed",
        "count": 1
    },
    {
        "name": "Test",
        "status": "In Progress",
        "count": 85
    },
    {
        "name": "Test",
        "status": "Failed",
        "count": 40
    },
    {
        "name": "Test",
        "status": "Active",
        "count": 1
    },
    {
        "name": "Test",
        "status": "Success",
        "count": 218
    },
    {
        "name": "Test 2",
        "status": "Active",
        "count": 1
    }
]

我想像这样格式化上面的最终图表数据,以便在图表中显示它:

[
    "labels": ['Test', 'Test 2', 'Test 3'],
    "data": [
        {
            name: 'Active',
            data: [1, 1, 1]
        },
        {
            name: 'Failed',
            data: [40, 1, 0]
        },
        {
            name: 'Success',
            data: [218, 0, 0]
        },
        {
            name: 'In Progress',
            data: [85, 0, 0]
        }
    ]
]

我正在尝试以这种方式格式化数据,但我做不到。是否有任何内置函数可用于使数据格式正确?

response = [
    {
        'labels': [],
        'data': [],
    }
]
for row in data:
    if row['name'] not in response[0]['labels']:
        response[0]['labels'].append(row['name'])
        innerData = []
        for status in ['Active', 'Failed', 'Success', 'In Progress']:
            if status in row['status']:
                innerData.append(row['count'])
            else:
                innerData.append(0)
        response[0]['data'].append(
            {
                'name': status,
                'data': innerData,
            }
        )

为了获得唯一的标签,set comprehension can be used. Then you can use defaultdictkey:value 对,因为每个状态是键名和计数是对应的值。

from collections import defaultdict

data = [{"name": "Test 3", "status": "Active", "count": 1}, {"name": "Test 2", "status": "Failed", "count": 1}, {"name": "Test", "status": "In Progress", "count": 85},{"name": "Test", "status": "Failed", "count": 40},{"name": "Test", "status": "Active", "count": 1},{"name": "Test", "status": "Success", "count": 218}, {"name": "Test 2", "status": "Active", "count": 1}]

d = defaultdict(list)
labels = sorted({each['name'] for each in data})

for each in data:
    d[each['status']].append(each['count'])
# -> defaultdict(<class 'list'>, {'Active': [1, 1, 1], 'Failed': [1, 40], 'In Progress': [85], 'Success': [218]})

final_result = {'labels': labels, 'data': []}
for k, v in d.items():
    final_result['data'].append({'name': k, 'data': v})

final_result 看起来像:

{'labels': ['Test', 'Test 2', 'Test 3'], 'data': [{'name': 'Active', 'data': [1, 1, 1]}, {'name': 'Failed', 'data': [1, 40]}, {'name': 'In Progress', 'data': [85]}, {'name': 'Success', 'data': [218]}]}

为了用零填充小型列表,请参阅

基于 Rustam 的回答:

  1. 添加字典label_to_idx 用于按标签确定索引。
  2. lambda 函数传递给 defaultdict 以初始化正确大小和默认值的列表。
  3. 为每个状态填充 d,使用 label_to_idx 找到正确的索引以填充计数。
from collections import defaultdict

data = [
    {"name": "Test 3", "status": "Active",      "count": 1},
    {"name": "Test 2", "status": "Failed",      "count": 1},
    {"name": "Test",   "status": "In Progress", "count": 85},
    {"name": "Test",   "status": "Failed",      "count": 40},
    {"name": "Test",   "status": "Active",      "count": 1},
    {"name": "Test",   "status": "Success",     "count": 218},
    {"name": "Test 2", "status": "Active",      "count": 1},
]

labels = sorted({each['name'] for each in data})
label_to_idx = {label: idx for idx, label in enumerate(labels)}    # Added
d = defaultdict(lambda: [0] * len(labels))                         # Modified

for each in data:
    d[each['status']][label_to_idx[each['name']]] = each['count']  # Modified

final_result = {'labels': labels, 'data': []}
for k, v in d.items():
    final_result['data'].append({'name': k, 'data': v})
from pprint import pprint as pp

pp(final_result)
# {'data': [{'data': [1, 1, 1], 'name': 'Active'},
#           {'data': [40, 1, 0], 'name': 'Failed'},
#           {'data': [85, 0, 0], 'name': 'In Progress'},
#           {'data': [218, 0, 0], 'name': 'Success'}],
#  'labels': ['Test', 'Test 2', 'Test 3']}