Create dictionary with lambda (incremental counter inside)

Question

I'm newbie to python. I have this program:

wordlist = ['pea', 'rpai', 'rpai', 'schiai', 'pea', 'rpe', 'zoi', 'zoi', 'briai', 'rpe']
dictionary = {}
counter = 0
result = list(map(lambda x: dictionary[wordlist[x]] = dictionary.get(wordlist[x], counter +=1), wordlist))
print(result)

Result has to be:

result = [0, 1, 1, 2, 0, 3, 4, 4, 5, 3]

What I have to do is append all of the element in the list (as key) inside the dictionary with an incremental counter as value of the key. With this code I get "lambda cannot contain assignment. How can I do this? Thank you!

EDIT FOR EXPLANATION:

With the list of strings I have to create a dictionary with element of list of str as "argument" and value as "key"

The value is calculated like this: The first element of the list is 0. The following element, if it is a new string never appeared (unique) has last value (in this case 0) =+1. Instead if the new element is a duplicate string (there is already one in the dictionary) it take the same originale value as the first.

The dictionary will be:

{'pea': 0, 'rpai': 1, 'rpai': 1, 'schiai': 2, 'pea': 0, 'rpe': 3, 
 'zoi': 4, 'zoi': 4, 'briai': 5,'rpe': 3}

And result instead with list will be:

[0, 1, 1, 2, 0, 3, 4, 4, 5, 3]

Write a normal `for` loop and a normal function instead of using `map` and `lambda`. — mkrieger1, Nov 07 '20 at 21:54
I really didn't understand what the rule is for the final result — Henrique Hott, Nov 07 '20 at 22:05
Edited! Sorry for my bad explanation, these are my first post. Thanks for your patience! — Alessio, Nov 07 '20 at 22:35
If it's relevant, I wrote [an answer](https://stackoverflow.com/a/61945553/4518341) about why not to use `map(lambda)` and what to use instead. — wjandrea, Nov 07 '20 at 23:25

Marat · Answer 1 · 2020-11-07T22:59:16.087

2

I guess the easiest solution with vanilla Python is to use defaultdict:

from collections import defaultdict

wordlist = ["pea","rpai","rpai","schiai","pea","rpe", "zoi","zoi","briai","rpe"]
vocab = defaultdict(lambda: len(vocab))
# result will be [0, 1, 1, 2, 0, 3, 4, 4, 5, 3]
result = [vocab[word] for word in wordlist]

A more verbose equivalent, leading to the same result:

vocab = {}
result = []
for word in wordlist:
    if word not in vocab:
        vocab[word] = len(vocab)
    result.append(vocab[word])

edited Nov 07 '20 at 22:59

answered Nov 07 '20 at 22:50

Marat

15,215
2
39
48

Without import, is there a term to call a python code without any import? Thanks for your help – Alessio Nov 07 '20 at 22:54
actually, @Frank's `.setdefault` version looks cleaner – Marat Nov 07 '20 at 23:10

Frank · Accepted Answer · 2020-11-07T23:07:00.667

1

Update: Use dictionary's setdefault then.

wordlist = ["pea","rpai","rpai","schiai","pea","rpe", "zoi","zoi","briai","rpe"]
dic = {}
res = list(map(lambda x: dic.setdefault(x, len(dic)), wordlist))
print(res)

Dictonary can't have same keys. You just need a for loop:

wordlist = ["pea","rpai","rpai","schiai","pea","rpe", "zoi","zoi","briai","rpe"]
c = 0
dic = {}
res = []
for i in range(len(wordlist)):
    word = wordlist[i]
    if word in dic:
        res.append(dic[word])
    else:
        dic[word] = c
        res.append(c)
        c += 1

print(res)

edited Nov 07 '20 at 23:07

answered Nov 07 '20 at 22:44

Frank

1,215
12
24

Since `i` isn't used in the logic, you can simplify to `for word in wordlist` – wjandrea Nov 07 '20 at 23:03

score 1 · Answer 3 · answered Nov 07 '20 at 22:59

1

Once you have the dictionary built the code for the lambda will be as follows.

list(map(lambda x: dictionary[x], wordlist))

This assumes you already have the keys and values of the dictionary populated. Is this the case, like so?

{'pea': 0, 'rpai': 1, 'schiai': 2, 'rpe': 3, 'zoi': 4, 'briai': 5, 'rpei': 6}

answered Nov 07 '20 at 22:59

Stu

421
3
7

It's much cleaner as a comprehension: `[dictionary[x] for x in wordlist]` – wjandrea Nov 07 '20 at 23:01
@wjandrea I know, but the OP has stated that the solution must use map and lambda – Stu Nov 07 '20 at 23:02
Sorry , I made a mistake with the request and the result is a little bit different, what i have to do, create a new post (and delete this) or append the edit? Thank you – Alessio Nov 08 '20 at 21:53

score 1 · Answer 4 · answered Nov 07 '20 at 22:59

All you need to do is make a dict with the first occurrence of each word, then look up each word in the dict. Don't use map and lambda, they'll only make it harder, or at least less readable.

first_occ = {}
counter = 0
for word in wordlist:
    if word not in first_occ:
        first_occ[word] = counter
        counter += 1

result = [first_occ[w] for w in wordlist]
print(result)  # -> [0, 1, 1, 2, 0, 3, 4, 4, 5, 3]

Create dictionary with lambda (incremental counter inside)

4 Answers4