迫于逻辑实在理不清了， Python 求助

{
"child": [{
"f_brandname": "奔驰",
"f_pyfirstletter": "B",
"seriesname": "载货车"
}, {
"f_brandname": "北奔重卡",
"f_pyfirstletter": "B",
"seriesname": "载货车"
}, {
"f_brandname": "福田欧曼",
"f_pyfirstletter": "F",
"seriesname": "载货车"
}, {
"f_brandname": "江铃汽车",
"f_pyfirstletter": "J",
"seriesname": "载货车"
}]
}

变成

{
"child": {
"B": [{
"f_brandname": "奔驰",
"f_pyfirstletter": "B",
"seriesname": "载货车"
}, {
"f_brandname": "北奔重卡",
"f_pyfirstletter": "B",
"seriesname": "载货车"
}],
"F": [{
"f_brandname": "福田欧曼",
"f_pyfirstletter": "F",
"seriesname": "载货车"
}],
"J": [{
"f_brandname": "江铃汽车",
"f_pyfirstletter": "J",
"seriesname": "载货车"
}]
}
}

就是把每个小字典给归类了，还有好多个品牌，这个是少数
语言是 python，求 demo，

seriesname

货车

f_brandname

f_pyfirstletter

14 条回复 2021-01-17 00:05:15 +08:00

taogen

2021-01-16 23:21:31 +08:00 via Android

没写过 Python 。不过把对象数组变成哈希数组，不会很难吧

yixiugegegege

2021-01-16 23:27:52 +08:00

@taogen 我百度一手，谢谢

nvkou

2021-01-16 23:28:36 +08:00 via Android

Java 的话第一反应是 stream. map. groupingby
Python 的话我估计也会这么做
大不了先拿 firstletter 的集合，再用这个集合过滤到不同组。也就是先建立键，再通过键过滤值

eggshell

2021-01-16 23:34:55 +08:00

用 itertools.groupby

xiaoming1992

2021-01-16 23:40:46 +08:00 via Android

就用 for in 也行吧

wzwwzw

2021-01-16 23:43:55 +08:00

from itertools import groupby

before_data = {
"child": [
{"f_brandname": "奔驰", "f_pyfirstletter": "B", "seriesname": "载货车"},
{"f_brandname": "北奔重卡", "f_pyfirstletter": "B", "seriesname": "载货车"},
{"f_brandname": "福田欧曼", "f_pyfirstletter": "F", "seriesname": "载货车"},
{"f_brandname": "江铃汽车", "f_pyfirstletter": "J", "seriesname": "载货车"},
]
}

after_data = {"child": {}}
for j, i in groupby(before_data["child"], key=lambda x: x["f_pyfirstletter"]):
if j not in after_data["child"]:
after_data_example["child"][j] = []
for k in i:
after_data_example["child"][j].append(k)

print(after_data_example)

zyx199199

2021-01-16 23:46:16 +08:00

from itertools import groupby

{k: list(v) for k,v in groupby(data['child'], key=lambda x: x["f_pyfirstletter"])}
```

yixiugegegege

2021-01-16 23:46:28 +08:00

@eggshell 卧槽，解决了，老哥，谢谢！！！

my8100

2021-01-16 23:46:43 +08:00 via iPhone

from collections import defaultdict

child_dict = defaultdict(list)
for d in data["child"]:
child_dict[d["f_pyfirstletter"]].append(d)

assert {"child": child_dict} == target_data

yixiugegegege

2021-01-16 23:46:55 +08:00

@wzwwzw 谢谢老哥的代码，终于解决了哭

yixiugegegege

2021-01-16 23:48:18 +08:00

@zyx199199 谢谢老哥解决了，喜极而泣，祝福本月无 bug，无迫于

yixiugegegege

2021-01-16 23:49:05 +08:00

@my8100 谢谢老哥，你，爱你！

imn1

2021-01-16 23:58:51 +08:00

new = {'child': {x["f_pyfirstletter"]: [] for x in d["child"]}}
[new['child'][x["f_pyfirstletter"]].append(x) for x in d["child"]]

下划线为空格
d 为源字典，new 为结果

imn1

2021-01-17 00:05:15 +08:00

# 13 “下划线为空格”不需要
原来是用 for 的，有缩进，后来改成列表表达式，不需要缩进了