Question

I am currently struggling with an assignment. The solution would input a txt file and run through counting the number of palindromes and their frequency. I need to use Map reduce to create to do so

For example: the string “bab bab bab cab cac dad” would output:

bab 3
cab 1
dad 1

Here is what I have so far

def palindrome(string):
    palindromes = []
    for word in string.split(" "):
        if (word == word[::-1]):
            palindromes.append(word)
    return palindromes 

string = "abc abd bab tab cab tat yay uaefdfdu"
print map(lambda x: palindrome(x), ["bab abc dab bab bab dad crap pap pap "])

Currently prints

[['bab', 'bab', 'bab', 'dad', 'pap', 'pap', '']]

Here is my attempt so far at the reduce section

def p(lists):
for list in lists:

set_h = set(list) 

return set_h

with the p function I want to create a set of all palindromes found. Then run a count of the palindromes on the list and make a dict out of this

print reduce(p, [['bab', 'bab', 'bab', 'dad', 'pap', 'pap', '']])

Am I on the right track?

Answer 1

Split your string into a list before you map it. map() is for lists, sets, and dicts, not strings.

word_list = words_str.split(" ")

Avoid using map-filter-reduce unless your assignment dictates it; GVR says so. The proper solution uses Python’s list comprehension syntax. In fact, you can do it with a pretty nasty one-liner:

pal_count = {
    x: word_list.count(x)  # reduce-ish
    for x in word_list     # map-ish
    if x == x[::-1]        # filter-ish
    }
for x, y in pal_count.iteritems():
    print x, y             # print the results!

Breaking it down…

1. Catch this in a dictionary object to print it later: pal_count = {
2. Define the return objects: x: word_list.count(x) We use key:value syntax to associate the palindrome, x, with its number of occurrences. count() is like a built-in reduce function for lists.
3. Iterate through our list with a for loop, assigning the current value to ‘x’: for x in word_list
4. We only want to return palindromes, so we add a comparison operator to filter out bad values: if x == x[::-1] # cool logic, btw
5. Hurray! }

By the way, I’m only doing your homework because I never did mine.

The slower, less flexible, less portable, less awesome equivalent uses nested for loops:

pal_count = dict()
for x in word_list:                     # same loop
    if x == x[::-1]                     # is this a palindrome?
        if x in pal_count:              # have we seen before?
            pal_count[x] += 1
        else:                           # this one is new!
            pal_count.setdefault(x, 1)