Accumulating a Maximum Value¶
Now let’s take a detour for a moment, and see how we can use the accumulator pattern to compute the maximum value in a list. Then you’ll have an exercise to find the key in a dictionary that has the largest associated value.
To accumulate the maximum value in a list, you can have your accumulator variable keep track of the max-so-far. Initialize it to the first item in the list. Then iterate through the rest of the items. For each, if it’s bigger than the max-so-far, replace the accumulator variable’s value with the value of the current item.
Step through the execution of this code to get a feel for how it works.
(dict_accum_9)
Now, you may notice that this code will break if there isn’t more than one item in L. You would get an error on line 4 for trying to access item L[1], which is the second item. If we assume that L will have only numbers >= 0, we can initialize the max-so-far to be 0 and loop through all of the items in L.
(dict_accum_10)
We can do a similar thing with a dictionary to find the maximum value. You can loop through the keys and replace the max-so-far whenever the current key’s associated value is greater than the max-so-far.
Check your understanding
- I
- c will be bound to a key, which is a string; you can't compare that to a number.
- II
- That will treate the current value of a as a key in the dictionary and update that key's value. You want to update a instead.
- III
- When the value associated with the current key c is bigger than the max so far, replace the max so far with that value.
- IV
- That will set a to be the current key, a string like 'a', not a value like 194.
sort-instances-1: Which is the right code block to use in place of line 5 if we want to print out the maximum value?
d = {'a': 194, 'b': 54, 'c':34, 'd': 44, 'e': 312, 'full':31}
a = 0
for c in d:
# <what code goes here? See below options>
print("max value is " + a)
# I.
if c > a:
a = c
# II.
if d[c] > a:
d[a] = c
# III.
if d[c] > a:
a = d[c]
# IV.
if d[c] > a:
a = c