Return 列表的名称，如果它包含所需的值 (Python)

Question

我正在尝试获取包含所需值的列表（类型列表）的名称：

def f(Value):
    
    a = ['a','b','c']
    b = ['d','e','f']
    d = ['g','h','i']
    c = ['j','k','l']
    w = next(n for n,v in filter(lambda t: isinstance(t[1],list), locals().items()) if value in v)
    return(w)

f()

好吧，此代码将 return 列表的名称但键入字符串，因此我以后将无法使用它。提前谢谢你

Answer 1

您的代码中有几个错误 - 如果您想要列表，那么 return 它而不是它的名称：

def f(value):                 # capitalization not correct
    
    a = ['a','b','c']
    b = ['d','e','f']
    d = ['g','h','i']
    c = ['j','k','l']

    # set w to v not n - n is the local name, v is the value
    w = next(v for n,v in filter(lambda t: isinstance(t[1],list),
                                 locals().items()) if value in v)

    return w

# needs to be called with parameter
what_list = f("h")

print( what_list )

输出：

['g', 'h', 'i']

我不明白为什么要在这个最小的例子中使用 locals().items()，你可以这样做

for li in [a,b,c,d]:  # simply add more lists if needed, no need to inspect locals
    if value in li:
        return li

相反。

Return 列表的名称，如果它包含所需的值 (Python)

Return the name of a list if it contain the needed value (Python)

python

list