從dict中提取值的pythonic方法是什么

Question

從字典中提取值的最佳方法是什么？ 假設我們有一個字典列表：

projects = [{'project': 'project_name1', 
             'dst-repo': 'some_dst_path', 
             'src-repo': 'some_src_path', 
             'branches': ['*']},
            {...}, 
            {...}]

現在，我只是遍歷此字典並獲取值，例如：

   for project in projects:
       project_name = project.get('project')
       project_src = ....
       project_dst = ....
       ....
       ....

因此問題是：“是否還有更多的Python方法通過字典從鍵中提取值，而這些方法不允許為新變量賦值那么多行代碼？”

Answer 1

您所做的沒有什么錯，但是您可以通過使用列表推導從當前詞典中提取值來使其更緊湊。 例如，

projects = [
    {
        'project': 'project_name1', 
        'dst-repo': 'some_dst_path', 
        'src-repo': 'some_src_path', 
        'branches': ['*']
    },
]

keys = ['project', 'src-repo', 'dst-repo', 'branches']
for project in projects:
    name, src, dst, branches = [project[k] for k in keys]
    # Do stuff with the values
    print(name, src, dst, branches)

輸出

project_name1 some_src_path some_dst_path ['*']

但是，如果鍵的數量很大，此方法將變得笨拙。

---

如果dict中有時缺少鍵，那么您將需要使用.get方法，該方法將為丟失的鍵返回None （除非您將默認參數傳遞給它）：

name, src, dst, branches = [project.get(k) for k in keys]

如果每個鍵都需要特定的默認值，則可以將它們放入字典中，例如

defaults = {
    'project': 'NONAME',
    'src-repo': 'NOSRC',
    'dst-repo': 'NODEST',
    'branches': ['*'],
}

projects = [
    {
        'project': 'project_name1', 
        'src-repo': 'some_src_path', 
    },
]

keys = ['project', 'src-repo', 'dst-repo', 'branches']
for project in projects:
    name, src, dst, branches = [project.get(k, defaults[k]) for k in keys]
    # Do stuff with the values
    print(name, src, dst, branches)

輸出

project_name1 some_src_path NODEST ['*']

Answer 2

out = [elt.values() for elt in projects]

Answer 3

   for project in projects:
       project_name = project['project']
       project_src = ....
       project_dst = ....
       ....
       ....

我不確定您輸入的字數會減少

//編輯：好的，看來我誤解了這個問題：假設我們有一個像這樣的字典列表：

projects = [ {'project': "proj1", 'other': "value1", 'other2': "value2"},
             {'project': "proj2", 'other': "value3", 'other2': "value4"},
             {'project': "proj2", 'other': "value3", 'other2': "value4"} ]

要提取project字段的列表，可以使用以下表達式：

projects_names =  [x['project'] for x in projects]

這將遍歷項目列表，從每個字典中提取“項目”鍵的值。