If elif else alternative

[brocq_18]

What is a more Pythonic way of doing this? I've seen some examples using tuples and dictionaries but they don't seem to return the same varialbe.

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if integerLength == 1:     n = 1 elif integerLength == 2:     n = 10 elif integerLength == 3:         n = 100 elif integerLength == 4:     n = 1000 elif integerLength == 5:         n = 10000 else:     n = 100000

[Larz60+]

What is the context? What are you doing with it.

[buran]

two alternatives

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>>> integerLength = 3 >>> n = 100000 if integerLength > 5 else 10**integerLength # using ternary expression >>> n 1000 >>> n = 10**min(integerLength, 5) >>> n 1000

also, for ternary expressions check this

[brocq_18]

Code in full;

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import math   maxValue = 415.2  integerPart = str(maxValue).split('.')[0] # selects the integer part integerLength = len(integerPart)   if integerLength == 1:     n = 1 elif integerLength == 2:     n = 10 elif integerLength == 3:         n = 100 elif integerLength == 4:     n = 1000 elif integerLength == 5:         n = 10000 else:     n = 100000       maxValueAdd = float(maxValue) + n + 0.1 # n allows for a value to be returned that is higher by the same factor  d = int(math.ceil(maxValueAdd)) # calculates the upper bound and returns an integer   print "Maximum value = {}".format(maxValue) print "Y axis maximum = {}".format(d)

[metulburr]

What is a more Pythonic way of doing this? I've seen some examples using tuples and dictionaries

It would be something along the lines of.

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d = {     1:1,     2:10,     3:100,     4:1000,     5:10000 }   if integerLength in d:     n = d[integerLength] else:     n = 100000

Or if you wanted the else clause in the dict 

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d = {     0:100000,     1:1,     2:10,     3:100,     4:1000,     5:10000 }   if intergerLength in d:     n = d[interLength] else:     n = d[0]

dont quote me on that though, i literally just woke up and it takes a few for my brain to start 

[buran]

based on metulburr's suggestion:

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>>> d = {    1:1,    2:10,    3:100,    4:1000,    5:10000 } >>> n = d.get(integerLength, 100000)

---

in addition one could use dict comprehension to construct the dict

1	d = {k+1:10**k for k in range(5)}

[brocq_18]

1	d = {k+1:10**k for k in range(5)}

That's great, as potentially the range may be higher than the 100000 I've detailed. 

[ankit]

More pythonic way and i would say more easy way to implement this is using dictionary and list comprehension :

dict = {'
       1':1,
       '2':10,
       '3':100,
       '4':1000,
       '5':10000,
       'rest':100000,
       }

n=raw_input() # any input

print dict['rest'] if len([dict[i] for i in dict if i==n] )==0 else [dict[i] for i in dict if i==n][0] # your result

[wavic]

It's almost unreadable.

[buran]

@ankit - there is nothing pythonic in this

1	dict['rest'] if len([dict[i] for i in dict if i==n] )==0 else [dict[i] for i in dict if i==n][0]