How to replace all values (by a function) in nested list in Python?

Question

There are many suggestions but I can't make it stick. The closest solution I have below. The data is like:

my_list = ([1, [2, 3], 4, [5, [6 , [], [8, 9]], 10]], 8)

and a function

def convert(G, l, d):
    z = []
    if isinstance(x, list):
        print(x)
        z.append(convert(G, x, d))
    else:
        print(x)
        z.append([v[d] for n,v in G.nodes(data=True) if n == l])
    return z

execution:

print(convert(G, my_list, "name"))

It gives empty array but print(x) gets the source as is. I'm close I guess. The problem is I don't know how to pass l at if n == l as a integer not list.

EDIT

Outpit: (['a', ['b', 'c'], 'd', ['e', ['f' , [], ['g', 'h']], 'j']], 'g) in the same format (nested). The custom function just takes each element (int) and returns its attribute from dict (letter).

Another try:

my_list = ([1, [2, 3], 4, [5, [6 , [], [8, 9]], 10]], 8)
z = []
def convert(G, l, d):
    z.append([convert(G, x, d) if isinstance(x, list) else [v[d] for n,v in G.nodes(data=True) if n == x]])
    return z

z = convert(G, my_`list, "name")
print(z)

This part is a custom function:

[v[d] for n,v in G.nodes(data=True) if n == l]

so G can be whatever. You can take it as [...]

For example: G.nodes = {0: {'name': 'a'}, 1: {'name': 'b'}, 2: {'name': 'c'}, 3: {'name': 'd'}}

Can you make sure you've included an minimal reproducible example? For example, what's G? As @Peter.k mentioned, what do you expect the output to be? — Chris Larson
– Chris Larson, Commented Feb 8, 2019 at 16:11

tobias_k · Accepted Answer · 2019-02-08 16:23:18Z

5

It seems like you are looking for a kind of map function that works for nested lists. Try this:

def deepmap(f, lst):
    return [deepmap(f, x) if isinstance(x, list) else f(x) for x in lst]

This will convert the outer tuple to another list, though. If you need that to be a tuple, or if there can be more tuples within, it should not be too hard to extend that function accordingly. Example:

>>> my_list = ([1, [2, 3], 4, [5, [6 , [], [8, 9]], 10]], 8)
>>> deepmap(str, my_list)
[['1', ['2', '3'], '4', ['5', ['6', [], ['8', '9']], '10']], '8']

In your case, the mapping function f might be something like this, not entirely sure though. (You might also want to change your G.nodes data structure so you don't have to iterate all the items but can access item x directly.)

f = lambda x: next(v[d] for n,v in G.nodes(data=True) if n == x)

edited Feb 8, 2019 at 16:23

answered Feb 8, 2019 at 16:17

tobias_k

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1 Comment

Peter.k Over a year ago

Yes. This solution works as well. Just needed to pass d argument to deepmap and lambda the rest is as is.

Bob Zimmermann · Accepted Answer · 2019-02-08 16:28:04Z

1

A simplified version, following your structure, might look like this:

def v(x):
    return x + 1

def convert(l):
    z = []
    for x in l:
        if isinstance(x, list):
            z.append(convert(x))
        else:
            z.append(v(x))
    return z

This results in the expected nesting:

>>> convert([[1, [2, 3], 4, [5, [6 , [], [8, 9]], 10]], 8])
[[2, [3, 4], 5, [6, [7, [], [9, 10]], 11]], 9]

edited Feb 8, 2019 at 16:28

answered Feb 8, 2019 at 16:22

Bob Zimmermann

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1 Comment

Peter.k Over a year ago

Yes, thanks. This is oke. I forgot to iterate for x in l:.

adrtam · Accepted Answer · 2019-02-08 16:18:35Z

0

You just need to make a recursive function to do the convert:

my_list = ([1, [2, 3], 4, [5, [6 , [], [8, 9]], 10]], 8)
mapping = {1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i', 10: 'j'}

def convert(inlist):
    def transform(x):
        if isinstance(x, list):
            return convert(x)
        else:
            return mapping.get(x)
    return [transform(x) for x in inlist]
print(convert(my_list))

answered Feb 8, 2019 at 16:18

adrtam

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Comments

mad_ · Accepted Answer · 2019-02-08 16:24:27Z

0

Define a custom mapper

import string
def mapper(item):
    if not isinstance(item,list):
        return d[item]
    else:
        return [mapper(i) for i in item]

d=dict(zip(range(1, len(string.ascii_lowercase) + 1), string.ascii_lowercase))

Mapping dict will look like this

{1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i', 10: 'j', 11: 'k', 12: 'l', 13: 'm', 14: 'n', 15: 'o', 16: 'p', 17: 'q', 18: 'r', 19: 's', 20: 't', 21: 'u', 22: 'v', 23: 'w', 24: 'x', 25: 'y', 26: 'z'}

my_list = ([1, [2, 3], 4, [5, [6 , [], [8, 9]], 10]], 8)
tuple([mapper(i) for i in my_list])

You can also use map function with custom mapper

tuple(map(mapper,my_list))

Output

(['a', ['b', 'c'], 'd', ['e', ['f', [], ['h', 'i']], 'j']], 'h')

edited Feb 8, 2019 at 16:24

answered Feb 8, 2019 at 16:18

mad_

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