4

I want to update the keys of my dictionary c with its new keys k_new. Even though I am referring different stack overflow questions like this it does not get updated. Please tell me where I make it wrong.

from nltk.stem import WordNetLemmatizer
lemmatizer = WordNetLemmatizer()
c = {'apples': 3, 'biscuits and tea': 3, 'oranges and onions': 4}
for k in c:
    splits=k.split()
    k_new= " ".join(lemmatizer.lemmatize(w.lower()) for w in splits)
    c[k_new] = c.pop(k)
print(c)

PS: I also used:

c[k_new] = c[k]
del c[k]

Then I get RuntimeError: dictionary changed size during iteration

Please help me

Improve this question
2

2 Answers 2

Reset to default
3

You update the dictionary while you iterate over it:

from nltk.stem import WordNetLemmatizer
lemmatizer = WordNetLemmatizer()
c = {'apples': 3, 'biscuits and tea': 3, 'oranges and onions': 4}
for k in c:  # iterate over c
    splits=k.split()
    k_new= " ".join(lemmatizer.lemmatize(w.lower()) for w in splits)
    c[k_new] = c.pop(k)  # update (two times) c
print(c)

updating a collection while you iterate over it is usually a very bad idea. Most data structures are not designed to handle this.

You can however construct a new dictionary:

from nltk.stem import WordNetLemmatizer
lemmatizer = WordNetLemmatizer()
c = {'apples': 3, 'biscuits and tea': 3, 'oranges and onions': 4}
c_new = {}
for k in c:
    splits=k.split()
    k_new= " ".join(lemmatizer.lemmatize(w.lower()) for w in splits)
    c_new[k_new] = c[k]
print(c_new)

We can make this more elegant by using dictionary comprehension:

{" ".join(lemmatizer.lemmatize(w.lower()) for w in k.split()): v
 for k,v in c.items()}

this one-liner constructs a new dictionary where we iterate over key-value pairs k,v of c, and add a key " ".join(lemmatizer.lemmatize(w.lower()) for w in k.split()) that we associate with the value v.

Improve this answer
Sign up to request clarification or add additional context in comments.

5 Comments

@thebjorn: yes. That is what I do in the list comprehension: I propose a fix, but then aim to show an - imho - more elegant way to do it.
both your solutions create new dicts, which is fine but not strictly necessary in this case since the OP can mutate his dict in place if he iterates over c.items() (or creates any other separate data structure to hold the keys). Personally I think the dict comprehension is too complex in this case (and the : v almost disappears), but that's probably more of a taste issue..
@thebjorn: but then we will use memory to hold the keys. c.items() is an iterable as well, so this is not a solution either (gives the same error, in Python-3.x). Since items() is a proxy that iterates almost the same way over the dict, but takes the value in the dict as well into account.
that's a Py3 misfeature, in my world c.items() returns a list :-D ps: you're holding two entire copies of the dictionary in memory, so the c.items() (py2) or list(c)don't consume any more memory..
@thebjorn: then how is the list maintained? It means a malloc to construct a list object, that has at least as size the size of a reference pointer times the number of items in it. Since one want the ammortized cost of an append to be O(1), usually it can be as large as the next power of two.
2

Iterating over dictionary while it is changing might have weired effects since you are using a real-time reference to the keys. Just make a list out of the key view and it will work:

for k in list(c):
    ...
Improve this answer

1 Comment

Thank you for the answer. However, I want to get the values as well assigned for the keys while changing keys. Can we do it with lists?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.