NameError: global name '_' is not defined

Question

The code:

if (query_id, _) in hashtable[bucket]:

I expected this to work like in a for loop, but instead it gives this error:

NameError: global name '_' is not defined

hastable[bucket] is a list of pairs if that matters _{(which I doubt)}. Any ideas?

I am interested in pairs that have `query_id` as their first member and I am not worried for the second one @PadraicCunningham. — gsamaras, May 11 '16 at 09:26
Then unpacking in a for loop as suggested is what you want, that or looping and using `if sub[0] == ..` — Padraic Cunningham, May 11 '16 at 09:28
Yes @PadraicCunningham, that would work too, thank you! Oh, why a downvote? I searched before asking and I couldn't figure it out! — gsamaras, May 11 '16 at 09:28

phihag · Accepted Answer · 2016-05-11T09:26:00.670

7

The x in y is not magic; it's basically the same as y.__contains__(x). Therefore, in cannot search with placeholders; the left argument is fully evaluated. Instead, use

if any(query_id == qid for (qid, _) in hashtable[bucket]):

edited May 11 '16 at 09:26

answered May 11 '16 at 09:23

phihag

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OK that worked and I understood why my code failed. Can you please explain what you did? :) Moreover, do you think that my question is so bad to have a -1 score? :/ – gsamaras May 11 '16 at 09:28
`any` evaluates every element of its argument. The generator expression forming its argument yields True iff the first item matches `query_id`. – phihag May 11 '16 at 10:48

score 3 · Answer 2 · edited May 11 '16 at 09:34

3

In a for loop (like the one you linked) the variable called _ is defined. As far as I can tell, you didn't define it anywhere. What did you expect _ to represent?

_ is just a normal variable name in Python, except in the interactive interpreter (where it represents the last variable output).

edited May 11 '16 at 09:34

gsamaras

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answered May 11 '16 at 09:26

Matthias Schreiber

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That should be a comment Matthias. I would expect the 3rd item of the top answer I linked to. Moreover, I would expect that `[tag:Python]` defines it automatically. – gsamaras May 11 '16 at 09:27
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I am sorry, I updated it and hope that is better as an answer. – Matthias Schreiber May 11 '16 at 09:33

score 1 · Answer 3 · edited May 11 '16 at 10:27

1

In the concept where you use it only stores the last output variable. I'd go with:

if query_id in (x[0] for x in hashtable[bucket]):

edited May 11 '16 at 10:27

armatita

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answered May 11 '16 at 09:44

ádi

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NameError: global name '_' is not defined

3 Answers3