从列表指向字典变量

Question

假设你有一个清单

a = [3,4,1]

我想用这些信息来指向字典：

b[3][4][1]

现在，我需要的是例行公事。看到该值后，在 b 的位置内读取和写入一个值。

我不喜欢复制变量。我想直接改变变量b的内容。

Answer 1

假设

b

是一个嵌套字典，你可以这样做

reduce(dict.get, a, b)

访问

b[3][4][1]

。

对于更通用的对象类型，请使用

reduce(operator.getitem, a, b)

写入值有点复杂：

reduce(dict.get, a[:-1], b)[a[-1]] = new_value

所有这些都假设您现在不提前知道

a

中的元素数量。如果你这样做，你可以接受neves的回答。

Answer 2

这将是基本算法：

获取物品的价值：

mylist = [3, 4, 1]
current = mydict
for item in mylist:
    current = current[item]
print(current)

设置项目的值：

mylist = [3, 4, 1]
newvalue = "foo"

current = mydict
for item in mylist[:-1]:
    current = current[item]
current[mylist[-1]] = newvalue

Answer 3

假设列表长度是固定的并且已知

a = [3, 4, 1]
x, y, z = a
print b[x][y][z]

你可以把它放在一个函数中

Answer 4

让我们编写自己的

getnesteditem

和

setnesteditem

：

def getnesteditem(xs, path):
    for i in path:
        xs = xs[i]
    return xs

def setnesteditem(xs, path, newvalue):
    for i in path[:-1]:
        xs = xs[i]
    xs[path[-1]] = newvalue

并尝试使用您的索引列表

a = [3,4,1]

：

a = [3, 4, 1]
b = [[[15*i+3*j+k for k in range(3)] for j in range(5)] for i in range(4)]
print(*b, sep='\n')
# [[ 0,  1,  2], [ 3,  4,  5], [ 6,  7,  8], [ 9, 10, 11], [12, 13, 14]]
# [[15, 16, 17], [18, 19, 20], [21, 22, 23], [24, 25, 26], [27, 28, 29]]
# [[30, 31, 32], [33, 34, 35], [36, 37, 38], [39, 40, 41], [42, 43, 44]]
# [[45, 46, 47], [48, 49, 50], [51, 52, 53], [54, 55, 56], [57, 58, 59]]

print(getnesteditem(b, a))
# 58

setnesteditem(b, a, -9)
print(*b, sep='\n')
# [[ 0,  1,  2], [ 3,  4,  5], [ 6,  7,  8], [ 9, 10, 11], [12, 13, 14]]
# [[15, 16, 17], [18, 19, 20], [21, 22, 23], [24, 25, 26], [27, 28, 29]]
# [[30, 31, 32], [33, 34, 35], [36, 37, 38], [39, 40, 41], [42, 43, 44]]
# [[45, 46, 47], [48, 49, 50], [51, 52, 53], [54, 55, 56], [57, -9, 59]]