在Python 2.7中获取列表的长度作为字典中的值 [英] Obtaining length of list as a value in dictionary in Python 2.7

问题描述

我有两个列表和字典，如下所示:

I have two lists and dictionary as follows:

>>> var1=[1,2,3,4]
>>> var2=[5,6,7]
>>> dict={1:var1,2:var2}

我想从我的字典中找到可变元素的大小，即键值的长度.查找 help('dict')后，我只能找到返回键数的函数，即 dict .__ len __().我尝试了 Java 方法(希望它可以工作)即 len(dict.items()[0]) 但它评估为 2.

I want to find the size of the mutable element from my dictionary i.e. the length of the value for a key. After looking up the help('dict'), I could only find the function to return number of keys i.e. dict.__len__(). I tried the Java method(hoping that it could work) i.e. len(dict.items()[0]) but it evaluated to 2.

我打算找到这个:
第一个键的值的长度: 4
第二个键的值的长度: 3

当列表是词典的一部分时，而不是单独的列表(如果它们的长度为 len(list)).

when the lists are a part of the dictionary and not as individual lists in case their length is len(list).

任何建议都会有很大帮助.

Any suggestions will be of great help.

推荐答案

dict.items()是一个包含字典所有键/值元组的列表，例如:

dict.items() is a list containing all key/value-tuples of the dictionary, e.g.:

[(1, [1,2,3,4]), (2, [5,6,7])]

因此，如果您编写 len(dict.items()[0])，那么您会要求该项目列表的第一个元组的长度.由于字典的元组始终为2元组(对)，因此长度为 2 .如果您想要给定键的值的长度，请输入:

So if you write len(dict.items()[0]), then you ask for the length of the first tuple of that items-list. Since the tuples of dictionaries are always 2-tuples (pairs), you get the length 2. If you want the length of a value for a given key, then write:

len(dict[key])

Aso:尽量不要使用标准类型的名称(例如 str ， dict ， set 等)作为变量名称.Python不会抱怨，但是它会隐藏类型名称，并可能导致意外的行为.

Aso: Try not to use the names of standard types (like str, dict, set etc.) as variable names. Python does not complain, but it hides the type names and may result in unexpected behaviour.

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