PYTHON 3 (1) Dictionary Python upper case values present in a single key. (2) Why datatype sets in python does not return True element?

Question

(1) A snippet that takes a dictionary and uppercases every value in the dictionary and sets that value back into the dictionary at the same key e.g.

Given:

d = {'a': ['amber', 'ash'], 'b': ['bart', 'betty']}

Result:

{'a': ['AMBER', 'ASH'], 'b': ['BART', 'BETTY']}

(2) Why datatype SET does not return TRUE element when printed? Eg. {'hi', 1, True} returns only {'hi', 1}

For (1) I am using something like this:

 d = {'a': ['amber', 'ash'], 'b': ['bart', 'betty']} 
 d.update((k, v.upper()) for k, v in d.items())

For (1) I am using something like this, d = {'a': ['amber', 'ash'], 'b': ['bart', 'betty']} d.update((k, v.upper()) for k,v in d.items()) — Vaibhav Sinha, Sep 01 '17 at 15:44
@Jean-FrançoisFabre yes, I'm new to python basics, i tried doing it successfully for one value but what is to be done for two or more values linked to a single key — Vaibhav Sinha, Sep 01 '17 at 15:49
@VaibhavSinha next time don't comment with your attempt, [edit] your question instead (at first I didn't see it and downvoted!) — Jean-François Fabre, Sep 01 '17 at 15:53
also, it's better to ask 2 questions because both parts of your question are unrelated (and the second question is quite puzzling) — Jean-François Fabre, Sep 01 '17 at 16:00

score 3 · Answer 1 · answered Sep 01 '17 at 15:49

3

(1)

d2 = {key:[name.upper() for name in names] for key, names in d.items()}

(2)

That seems to be because True == 1 yields True, which is what the Set uses to check if the value added is already in the Set and therefore has to be ignored.

answered Sep 01 '17 at 15:49

Jeronimo

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@Jean-François spoon-feeding askers of homework dumps is far from "nice". Please only encourage behaviour that's helpful for Stack Overflow. – Andras Deak -- Слава Україні Sep 01 '17 at 16:30
1

@AndrasDeak it's not a homework dump. It's homework, but not dump. There _is_ an attempt (I even retracted my close vote). I admit that the question should be 2 questions, though. that's the one reproach we can do to that question. – Jean-François Fabre Sep 01 '17 at 16:32
@Jean-François you're right, I missed that. The question needs some work. Still, pure code lines as answers are suboptimal. – Andras Deak -- Слава Україні Sep 01 '17 at 17:04
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@AndrasDeak that's his first answer. Give the guy some slack (check 1 rep answers, almost all in the NAA queue). But you're right. For the explanations, see _my_ answer :) – Jean-François Fabre Sep 01 '17 at 17:05

score 1 · Answer 2 · edited Sep 01 '17 at 16:05

(1) Could be shorter, just in one line as shown by others answers, but here is trade off between complexity and "Pythonicity":

d = {'a': ['amber', 'ash'], 'b': ['bart', 'betty']}
for k in d:
    d[k] = [i.upper() for i in d[k]]

print(d)

OUTPUT:

{'a': ['AMBER', 'ASH'], 'b': ['BART', 'BETTY']}

(2) Because True == 1 is true and Python set objects just have items that are differences between them.

Jean-François Fabre · Answer 3 · 2017-09-01T16:02:07.450

1

Your attempt is:

d.update((k, v.upper()) for k,v in d.items())

That doesn't work that way. For instance, v is a list, you cannot upper a list...

This kind of transformation is better done using a dictionary comprehension to rebuild a new version of d. You can do the upper part for each value using list comprehension:

d = {k:[v.upper() for v in vl] for k,vl in d.items()}

For your second question: since 1==True, the set keeps only the first inserted, which here is 1. but could have been True: example:

>>> {True,1}
{1}
>>> {True,1,True}
{True}
>>> {1,True}
{True}
>>>

more deterministic: pass a list to build the set instead of using the set notation:

>>> set([True,1])
{True}
>>> set([1,True])
{1}

edited Sep 01 '17 at 16:02

answered Sep 01 '17 at 15:52

Jean-François Fabre

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Thank you for your time! I get your point now My snippet fails because it is trying to change the case of the list which makes no sense. Thank you! – Vaibhav Sinha Sep 01 '17 at 15:54
Jeronimo answer is better because it uses meaningful names. The values are names. Once you wrote that, you know that you have to iterate on them. – Jean-François Fabre Sep 01 '17 at 15:55
"the `set` keeps only the first inserted"- I'm not sure that's right? – Chris_Rands Sep 01 '17 at 15:55
@Chris_Rands for `set(['hi',1,True])` I'm sure of this. But ok, let me dig into this (maybe matter for another question ... or duplicate :)) what the hell, it doesn't work :) – Jean-François Fabre Sep 01 '17 at 15:56
@Jean-FrançoisFabre But `{'hi',1,True}` is different (at least on my system on Py3) – Chris_Rands Sep 01 '17 at 15:58
@Chris_Rands >>> set(['hi',1,True]) {1, 'hi'} >>> set(['hi',True,1]) {True, 'hi'} >>> set(['hi',True,0]) {0, True, 'hi'} >>> – Rolf of Saxony Sep 01 '17 at 15:58
@Chris_Rands I stand corrected on "first inserted". It's just that you cannot predict the first inserted :) – Jean-François Fabre Sep 01 '17 at 15:59
@Jean-FrançoisFabre Isn't it the "last inserted" that is kept for `set` literals? Or am I misunderstanding you? – Chris_Rands Sep 01 '17 at 16:01
@Chris_Rands you may be right, but it's probably implementation defined. tested on py2 & 3 and it seems to be the same. I – Jean-François Fabre Sep 01 '17 at 16:02
@Jean-FrançoisFabre I might be wrong, and i also don't know if it's guaranteed. anyway, sorry I'm drifting off-topic... – Chris_Rands Sep 01 '17 at 16:07
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don't be. It's kind of interesting – Jean-François Fabre Sep 01 '17 at 16:07
@Chris_Rands https://stackoverflow.com/questions/46004328/order-of-insertion-in-sets :) in case you're not checking the questions now – Jean-François Fabre Sep 01 '17 at 16:19
@Jean-FrançoisFabre Ah thanks, I'll follow the discussion with interest! – Chris_Rands Sep 01 '17 at 17:49

PYTHON 3 (1) Dictionary Python upper case values present in a single key. (2) Why datatype sets in python does not return True element?

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