Question

我有一些......

[{"countryname": "Republic of Tunisia", "project_name": "TN: DTF Social Protection Reforms Support", "lendprojectcost": 5700000}, 
{"countryname": "Republic of Tunisia", "project_name": "Tunisia: Ecotourism and Conservation of Desert Biodiversity", "lendprojectcost": 9050000},
{"countryname": "Republic of Tunisia", "project_name": "Tunisia - Communications for policy reforms", "lendprojectcost": 600000},
{"countryname": "Republic of Tunisia", "project_name": "Tunisia - Governance, Opportunities and Jobs DPL", "lendprojectcost": 500000000}]

我想：

[{"countryname": "Republic of Tunisia", "project_name":all projects, "lendprojectcost":sum(..) }]

我怎么能这样做？我使用了Python / Flask / MongoDB。

Answer 1

作为熊猫极客，我会选择一个熊猫解决方案：从原始数据创建一个数据框（让我们称之为1: ~/.vimrc 2: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/getscriptPlugin.vim 3: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/gzip.vim 4: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/logiPat.vim 5: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/manpager.vim 6: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/matchparen.vim 7: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/netrwPlugin.vim 8: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/rrhelper.vim 9: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/spellfile.vim 10: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/tarPlugin.vim 11: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/tohtml.vim 12: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/vimballPlugin.vim 13: /usr/local/Cellar/vim/8.0.0134/share/vim/vim80/plugin/zipPlugin.vim），做所有必要的聚合，并构建另一个字典：

raw

Answer 2

您可以首先使用collections.defaultdict创建dict对象，以便为每个"countryname"创建唯一条目。然后将dict转换为您想要的回复。

例如：

from collections import defaultdict

country_project, country_sum = defaultdict(list), defaultdict(int)

for country in country_list:
    country_name = country["countryname"]
    country_project[country_name].append(country['project_name'])
    country_sum[country_name] += country['lendprojectcost']

# Map the `dict` to get desired result using list comprehension
new_list = [ {
        'countryname': country,
        'project_name': country_project[country],
        'lendprojectcost': country_sum[country]} for country in country_project]

# OR, via using plain `for` loop as:
# new_list = []
# for country in country_project:
#     new_list.append({
#         'countryname': country,
#         'project_name': country_project[country],
#         'lendprojectcost': country_sum[country]})

其中country_list是问题中提到的原始列表，而my_list保留的最终值是所需的结果：

[{
    'countryname': 'Republic of Tunisia', 
    'project_name': [
        'TN: DTF Social Protection Reforms Support', 
        'Tunisia: Ecotourism and Conservation of Desert Biodiversity', 
        'Tunisia - Communications for policy reforms', 
        'Tunisia - Governance, Opportunities and Jobs DPL'
    ], 
    'lendprojectcost': 515350000
}]

Answer 3

这里的其他答案都很棒，但这似乎是reduce的自然问题。您有一个列表，并且您正在尝试将列表中的元素与操作组合在一起，以便将列表大小“减少”到一个元素：

def update(x, y):
    x["project_name"].append(y["project_name"])
    x["lendprojectcost"] += y["lendprojectcost"]
    return x

result = [reduce(update, list_of_dicts, {"countryname": "Republic of Tunisia",
                                         "project_name": [],
                                         "lendprojectcost": 0})]

奖金功能：无需导入

遇到python问题

3 个答案: