On 07/09/2013 04:04 PM, HRK wrote:
Hello,
I'm pretty new to Python so while I can make some basic stuff work, it doesn't
look particularly pretty. I wanted to find better ways to solve the following
two problems (the current code works, but is clunky.) Any tips or help
appreciated. Thanks!
In your subject line you say "efficient" but here you say "(less)
clunky". Which is it?
For the first assignment, it's almost as simple as it's going to get,
though you could eliminate the if test by using the dict.get() method.
http://docs.python.org/2/library/stdtypes.html?highlight=dict.get#dict.get
PROBLEM #1
heights = {"Mount Everest": 8848, "Mount Cook": 3754, "K2": 8611,
"Ben Nevis": 1344, "Hekla": 1488}
def mountain_height(heights, mountain):
'''Return the height of the mountain.
If the mountain isn't in the dictionary, the
height returned should be -1.
The search should ignore any spaces before or after the name.
>>> mountain_height({'K2':8611, 'Hekla':1488}, "K2")
8611
>>> mountain_height({'K2':8611, 'Hekla':1488}, "Hekla ")
1488
>>> mountain_height({'K2':8611, 'Hekla':1488}, "Port Hills")
-1
'''
if mountain.strip() in heights:
return heights[mountain.strip()]
else:
return -1
print(mountain_height({'K2':8611, 'Hekla':1488}, "K2"))
#8611
print(mountain_height({'K2':8611, 'Hekla':1488}, "Hekla "))
#1488
print(mountain_height({'K2':8611, 'Hekla':1488}, "Port Hills"))
#-1
PROBLEM #2
def is_up_down(values):
'''Return True if values is an up-down-sequence, otherwise
return False.
A sequence [x[0], x[1], x[2], ..., x[n]] is an up-down-sequence
if there is an index k such that
x[0] <= x[1] <= x[2] <= ... <= x[k-1] <= x[k] and
x[k] >= x[k+1] >= x[k+2] >= ... >= x[n-1] >= x[n] hold.
That is, the first part of the sequence is increasing and the
second part is decreasing.
Either part may be empty, so any increasing sequence and any
decreasing sequence is an up-down-sequence.
'''
flip = 0
value = 0
for item in values:
if flip == 2:
if item > value:
return False
elif flip == 1:
if item < value:
flip += 1
value = item
elif flip == 0:
Usually the lst elif should be spelled "else". Or you should add
another else afterwards. Or add an assert() to double-check your logic.
if item > value:
flip += 1
value = item
return True
print(is_up_down([2,5,5,7,9,9,8]))
#True
print(is_up_down([2,5,5,7,9,8,9]))
#False
print(is_up_down([9,8]))
#True
print(is_up_down([13]))
#True
Here, your code may be about as fast as it's going to get. But it's
very hard to follow, which may be what you mean by 'clunky.' If it were
my problem, I'd make the code much slower by defining a second function
that checks whether a sequence is monotonic, and if so which way. Then
the main function will simply loop through the sequence, checking the
head and tail with the second function. As soon as it reaches a point
where the head and tail are both monotonic, and with the opposite
direction, you're done.
--
DaveA
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