efficient way to write 'or' statement

Question

From my standard input I take two words and I want to process them if they occur in a dictionary. This means that I can have both, one or none of the words match (in which case I will wrap up the code with an error message). This is very similar to a boolean OR gate. I can write my script like this:

if word1 in dict and word2 in dict:
    # process word1
    # process word2
elif word1 in dict:
    # process word1
elif word2 in dict:
    # process word2
else:
    # error msg

However, this seems rather redundant. Is there a simple alternative to this? I have a separate function that processes one of these words at a time.

`if word1 and word2 in dict` does not do what you think it does. It returns `True` for any `word1` that is not empty - also do not name variables after built ins, you shadow them `(list,dict,set,abs,min,max,tuple,...)` — Patrick Artner, Oct 06 '18 at 10:04
@PatrickArtner I am aware of the error, I meant to write it like this: `if word1 in dict and word2 in dict`. Thanks for notifying me though! And the `dict` variable name is purely for the example. — Hylke, Oct 06 '18 at 10:12
The value of the dictionary key is used while processing the word. — Hylke, Oct 06 '18 at 10:12
Please add somd dummy method calls for processing which make clear how those differ, especially if when both words are there, the processing is just the sequential individual processing. — Jeronimo, Oct 06 '18 at 10:27
@HylkevanderVeen Under the first if condition you should probably write #process word1 and another line #process word2. This would make your question clearer. — Ozair Kafray, Oct 06 '18 at 10:33

Ozair Kafray · Answer 1 · 2018-10-06T10:25:16.467

1

If processing word1 and word2 is independent of each other, then you can modify your code as follows:

word_found = False
if word1 in dict:
    #process word1
    word_found = True
if word2 in dict:
    #process word2
    word_found = True
if not word_found:
    #error message

edited Oct 06 '18 at 10:25

answered Oct 06 '18 at 10:18

Ozair Kafray

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As he said, "Mind you, I have a separate function which processes one of these words at a time.". So if both words are there, processing is apparently different from just doing both sequentially. Although, the question is a bit unclear about that.. – Jeronimo Oct 06 '18 at 10:19
1

Sorry I might not have written that clearly, I will make an edit. I meant to say I have a function which processes one word. This means when I want to process both words, I have to call the function twice. – Hylke Oct 06 '18 at 10:21
1

Generally you'd write `if not wordFound:` (or `word_found`); encouraging equality comparison like this can lead to unexpected results (e.g. https://stackoverflow.com/q/7479808/3001761). – jonrsharpe Oct 06 '18 at 10:22
@jonrsharpe Thanks. Corrected. I am a technology agnostic programmer. So not aware of code styling in python. – Ozair Kafray Oct 06 '18 at 10:26

Jeronimo · Accepted Answer · 2018-10-06T10:55:54.917

1

Now that it's clear that processing of both words is nothing special, i.e. just sequential processing of the individual words, you can reduce your code by moving up the error check to the top:

if word1 not in dict and word2 not in dict:
    # raise Exception or print and return
if word1 in dict:
    # process word1
if word2 in dict:
    # process word2

Readable and no extra state variables needed.

edited Oct 06 '18 at 10:55

answered Oct 06 '18 at 10:49

Jeronimo

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Patrick Artner · Answer 3 · 2018-10-06T10:52:00.513

You can leverage all(..) to check if multiple keys are in your dictionary - this will test all things until a False one is found or all are True:

def doSmthWithAllOfThem(words,dictionary):
    print(words, "are in it")
    print(sum(dictionary[word] for word in words) )

d = {"w{}".format(i):i for i in range(10)}

words = {"w1","w9","w4"} 

if all(w in d for w in words):
    doSmthWithAllOfThem(words,d)
elif "w9" in dict:
    pass
else:
    raise ValueError("not in")

Output:

{'w4', 'w9', 'w1'} are in it
14

You could also use set-intersection operation:

words = {"w1","w9","w4"} 

if words & set(d.keys()) == words:
    doSmthWithAllOfThem(words,d)

or (cudos to @jpp) the use set-issubset operation:

if words <= set(d.keys()):
    doSmthWithAllOfThem(words,d)

(Output in both set-usage cases identical to the one above)

The intersection returns only things that are in both sets - if the result afterwards is euqal to words all elements were in your dict.

efficient way to write 'or' statement

3 Answers3