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使用python检查字典列表中的单个元素_Python_List_Dictionary - Fatal编程技术网

使用python检查字典列表中的单个元素

使用python检查字典列表中的单个元素,python,list,dictionary,Python,List,Dictionary,您有这样一个字典列表: groups = {"1":["Susanne", "Laura", "Anna"], "2":["Kim", "Lisa", "Kevin"]; "3": ["Donald", "Hans", "Jack"], } 如何检查单个元素,例如:有多少名称以“A”开头 我什么都试过了,但总是出错…您可以将内部列表展平,然后检查列表中满足条件的列表数: sum(i.startswith('A') for i in [s for l in groups.

您有这样一个字典列表:

groups = {"1":["Susanne", "Laura", "Anna"],
    "2":["Kim", "Lisa", "Kevin"];
    "3": ["Donald", "Hans", "Jack"],
    }
如何检查单个元素,例如:有多少名称以“A”开头
我什么都试过了,但总是出错…

您可以将内部列表展平,然后检查列表中满足条件的列表数:

sum(i.startswith('A') for i in [s for l in groups.values() for s in l])
# 1

上述内容相当于以下for循环:

c = 0
for l in groups.values():
    for s in l:
           if s.startswith('A'):
                c += 1
你可以试试这个

sum(map(lambda x:x.startswith('A'),sum(groups.values(),[])))
#1

你尝试了什么?邮政编码,我们会帮你修复,但我们不会帮你。嘿,泰来回答。我需要一个不导入任何内容的解决方案,我认为这应该是可能的,但我不知道如何解决。(对python来说非常陌生)您还可以使用嵌套列表construction@M0NX(请参阅更新)将其展平。这个循环正是我搜索的:)