How to update values in nested dictionary if keys are in a list? [duplicate]

Question

Let's say i have a list of keys

key_lst = ["key1", "key2", "key3"]

and i have a value

value = "my_value"

and an example dict my_dict with this structure

{
"key1": {
    "key2": {
        "key3": "some_value"
        }
    },
}

How can I dynamically assign the new value in variable value to my_dict["key1"]["key2"]["key3"] by going thru / looping over my key_lst?

I can not just say my_dict["key1"]["key2"]["key3"] = value since the keys and the number of keys is changing. I always get the keys (the path that i have to save the value at) in a list...

The output I am looking for is {'key1': {'key2': {'key3': 'my_value'}}}. The dictionary structure is predefined.

I'm using Python 3.7

Is this really a duplicate? The code to answer is in the other thread, but conceptually I think the two questions are different. This is about setting a nested value while the link is focused on getting a nested value — Addison Klinke
– Addison Klinke, Commented Jun 15, 2022 at 19:32

jpp · Accepted Answer · 2019-01-11 11:13:44Z

17

Predefined dictionary structure: `functools.reduce`

You can define a function using functools.reduce to apply getitem repeatedly and then set a supplied value:

from functools import reduce
from operator import getitem

def set_nested_item(dataDict, mapList, val):
    """Set item in nested dictionary"""
    reduce(getitem, mapList[:-1], dataDict)[mapList[-1]] = val
    return dataDict

key_lst = ["key1", "key2", "key3"]
value = "my_value"
d = {"key1": {"key2": {"key3": "some_value"}}}

d = set_nested_item(d, key_lst, value)

print(d)
# {'key1': {'key2': {'key3': 'my_value'}}}

Note operator.getitem is used to access dict.__getitem__, or its more commonly used syntactic sugar dict[]. In this instance, functools.reduce calls getitem recursively on dataDict, successively using each value in mapList[:-1] as an argument. With [:-1], we intentionally leave out the last value, so we can use __setitem__ via dict[key] = value for the final key.

Arbitrary dictionary nesting: `collections.defaultdict`

If you wish to add items at arbitrary branches not yet been defined, you can construct a defaultdict. For this, you can first defaultify your regular dictionary input, then use set_nested_item as before:

from collections import defaultdict

def dd_rec():
    return defaultdict(dd_rec)

def defaultify(d):
    if not isinstance(d, dict):
        return d
    return defaultdict(dd_rec, {k: defaultify(v) for k, v in d.items()})

dd = defaultify(d)

key_lst = ["key1", "key2", "key5", "key6"]
value = "my_value2"
dd = set_nested_item(dd, key_lst, value)

print(dd)

# defaultdict(<function __main__.<lambda>>,
#             {'key1': defaultdict(<function __main__.<lambda>>,
#                          {'key2': defaultdict(<function __main__.<lambda>>,
#                                       {'key3': 'my_value',
#                                        'key5': defaultdict(<function __main__.<lambda>>,
#                                                    {'key6': 'my_value2'})})})})

edited Jan 11, 2019 at 11:13

answered Jan 10, 2019 at 23:05

jpp

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7 Comments

hqkhan Over a year ago

This is a pretty cool answer. Mind explaining a bit more on your set_nested_item function for me?

cs95 Over a year ago

But this assumes the particular structure already exists. What about setting new values? You will need a defaultdict, right? This cannot handle KeyErrors as it is.

jpp Over a year ago

@hqkhan, Added an explanation.

jpp Over a year ago

@coldspeed, Agreed, this doesn't handle KeyError. That said, not sure whether this is relevant for OP's requirements. For example, it's conceivable that the dictionary structure is given and you only want to set values; hence KeyError is what you want.

Endogen Over a year ago

JESUS CHRIST! That's a nice solution. Dict structure is actually given and i would expect a KeyError if the key is not in there. BTW: Not a requirement but this is more than three times faster than @coldspeed's answer. This is the one, thanks.

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cs95 · Accepted Answer · 2019-01-10 23:02:34Z

6

You can iteratively build/access levels using setdefault in a loop:

d = {}
d2 = d
for k in key_lst[:-1]:
    d2 = d2.setdefault(k, {})

d2[key_lst[-1]] = value
print(d)
# {'key1': {'key2': {'key3': 'my_value'}}}

d is the reference to your dictionary, and d2 is a throw-away reference that accesses inner levels at each iteration.

answered Jan 10, 2019 at 23:02

cs95

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1 Comment

BetaDev Over a year ago

Yes works perfectly. Awesome.

Julien · Accepted Answer · 2019-01-10 23:31:20Z

1

This is what you want:

def update(d, key_lst , val):
    for k in key_lst[:-1]:
        if k not in d:
            d[k] = {}
        d = d[k]
    d[key_lst[-1]] = val

d = {}

update(d, list('qwer'), 0)
# d = {'q': {'w': {'e': {'r': 0}}}}

You could use defaultdict too, it's neat in a sense but prints rather ugly...:

from collections import defaultdict

nest = lambda: defaultdict(nest)
d = nest()

def update(d, key_lst , val):
    for k in key_lst[:-1]:
        d = d[k]
    d[key_lst[-1]] = val

update(d, 'qwer', 0)

edited Jan 10, 2019 at 23:31

answered Jan 10, 2019 at 22:59

Julien

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Comments

mgc · Accepted Answer · 2019-01-10 23:12:09Z

0

I guess you can loop through your keys like this :

d = {}
a = d
for i in key_lst: 
    a[i] = {}
    if i == key_lst[-1]:
        a[i] = value
    else:
        a = a[i]
print(d)
# {'key1': {'key2': {'key3': 'my_value'}}}

Edit: I guess I misread the question and answered as if the dictionnary wasn't already existing. jpp answer is pretty neat otherwise I guess!

edited Jan 10, 2019 at 23:12

answered Jan 10, 2019 at 22:59

mgc

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Comments

Mehrdad Dowlatabadi · Accepted Answer · 2019-01-11 04:40:32Z

0

key_lst = ["key1", "key2", "key3"]
my_dict={
"key1": {
    "key2": {
        "key3": "some_value"
        }
    },
}

val=my_dict
#loop gets second to last key in chain(path) and assigns it to val
for x in key_lst[:-1]:
    val=val[x]
#now we can update value of last key, cause dictionary key is passed by reference
val[key_lst[-1]]="new value"

print (my_dict)

#{'key1': {'key2': {'key3': 'new value'}}}

edited Jan 11, 2019 at 4:40

answered Jan 10, 2019 at 23:20

Mehrdad Dowlatabadi

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6 Comments

Mehrdad Dowlatabadi Over a year ago

@Julien what is half broken?

Mehrdad Dowlatabadi Over a year ago

@Julien thats right thanks, but i can't see possibility of non existing key in question

Julien Over a year ago

I take that back. OP has clarified the dict structure is given in a comment to another answer... (wasn't the downvoter in the first place but upvoted to balance out :)

Mehrdad Dowlatabadi Over a year ago

@Julien thanks, i was thinking to delete my answer,you know i'm new and i was late to submit this answer and didn't know there's similar answer. next time i'll submit carefully :)

Kraang Prime Over a year ago

Please add some comments around it explaining why this is an answer, and/or what problem this code solves.

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