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.
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
What is the context? What are you doing with it.
two alternatives
>>> 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
Code in full;
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)
Quote: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.
d = {
1:1,
2:10,
3:100,
4:1000,
5:10000
}
if integerLength in d:
n = d[integerLength]
else:
n = 100000Or if you wanted the else clause in the dict
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
based on metulburr's suggestion:
>>> 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
d = {k+1:10**k for k in range(5)}
(Apr-11-2017, 12:07 PM)in addition one could use dict comprehension to construct the dict Wrote: [ -> ]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.
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
@
ankit - there is nothing pythonic in this
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]