有没有办法遍历列表并根据字典替换值？

Question

I have created a list of routes with the following layout:

routes = [[14,133,101,40,55,149,41,165,45,182,220,104,110,100,117,126,205,64,194,118,203],
 [7, 145, 111, 7, 180, 168, 136, 33, 70, 222, 190, 83, 233, 103],
 [185, 124, 185, 82, 195],
 [185, 91, 60, 8, 6],
 [220, 43, 179, 214],
 [7, 226, 187, 25, 152, 94, 46, 13, 19, 79, 125],
 [104, 72, 89, 51, 2, 172],
 [7, 147, 130, 160, 54, 116, 77, 156, 142, 78, 200, 122],
 [7, 175, 138, 49, 96, 148, 88, 123, 207, 97, 112, 169],
 [104, 201, 167, 53, 42, 15],
 [7, 95, 34, 137, 36, 20, 56, 164, 129, 5, 16],
 [3, 44, 71, 48, 102, 131, 139, 30, 221, 22, 57, 23, 66, 204, 99],
 [220, 114, 150, 217],
 [104, 170, 87, 174, 140, 134],
 [7, 193, 86, 202, 59, 143, 108, 21, 155, 198],
 [104, 162, 230, 166, 173],
 [185, 177, 127, 208, 158],
 [185, 227, 178, 176],
 [220, 224, 98],
 [185, 232, 37],
 [185, 225]]

These numbers correspond to a dictionary index, like the one below:

{1: 4,
 2: 38,
 3: 71,
 4: 90,
 5: 94,
 6: 101,
 7: 142,
 8: 163,
 9: 164,
 10: 196,
 ...
 234: 8360,
 235: 8507,
 236: 8545}

I want to iterate through the route list replacing every number with the corresponding value from the dictionary, eg convert the 7 in routes into 142. I have tried to do this using the code below:

([d.get(x,"No_key") for x in routes])

but this gives the following error:

TypeError: unhashable type: 'list'

It seems to work when I instead do this:

([d.get(x,"No_key") for x in routes[0]])

But I would like a way to carry this out for a route list of indeterminate length instead of having to go through the list manually row by row

Is there another way to iterate through and replace these values?

Answer 1

You have a list of lists, not a single flat list, so you have to use two nested list comprehensions:

routes = [
    [
        d.get(index, "No_key")
        for index in route
    ]
    for route in routes
]

Answer 2

Use 2 list comprehensions:

routes = [[2, 3, 1], [6, 5, 4]]
dct = {1: 4,
       2: 38,
       3: 71,
       4: 90,
       5: 94,
       6: 101,
}

routes = [[dct[k] for k in lst] for lst in routes]
print(routes)
# [[38, 71, 4], [101, 94, 90]]