I have code to sort a list of tuples:
s = "betty bought a bit of butter but the butter was bitter"
words = s.split()
l = []
k = []
unique_words = sorted(set(words))
for word in unique_words:
k.append(word)
l.append(words.count(word))
z = zip(k,l)
print z
reversed(sorted(z, key=lambda x: x[1]))
print z
z is the same, list doesn't get sorted or even reversed.
I am trying to sort by the integer value of count.
For an in-place sort, you should use z.sort().
If you insist on using sorted, then send the value back to z.
So, use either,
z.sort(key = lambda x:x[1])
z.reverse()
Or,
z = reversed(sorted(z, key=lambda x: x[1]))
Or, a more sophisticated solution could be:
z = sorted(z, key=lambda x: x[1], reverse= True)
As a matter of fact, you can get the end result more easily by using collections.Counter()
from collections import Counter
z = sorted(Counter(s.split()).items(), key = lambda x:x[1], reverse = True)
Sorting by two multiple keys are fine, you can pass them as a tuple. In your case, the solution would be:
# first sort by negatives of the second item, then alphabetically.
z = sorted(z, key=lambda x: (-x[1],x[0]))
Output:
[('butter', 2), ('a', 1), ('betty', 1), ('bit', 1), ('bitter', 1),
('bought', 1), ('but', 1), ('of', 1), ('the', 1), ('was', 1)]
python-2.7, but the errors you are getting imply Python3. Can you please clarify which python version you are using?reversed and sorted do not sort in-place; instead they return the newly sorted and reversed object. Change the second to last line to
z = list(reversed(sorted(z, key=lambda x: x[1])))
and it will work. The list call is because reversed returns an iterator rather than a list (on Python3, at least).
It might be a bit less verbose to do the following
z = sorted(z, key=lambda x: x[1], reverse=True)
It's almost correct - if you check help(reversed) in a Python REPL you'll find that it returns an iterator containing your sorted result based on your dict's values.
If you want z to store your updated, reversed sorted list on count, you'll need to reassign z:
z = list(reversed(sorted(z, key=lambda x: x[1])))
Edit: just to clarify, the outermost list conversion of the iterator objects 'converts' the iterator into a list of the objects contained inside the iterator.
Least changes to your code:
s = "betty bought a bit of butter but the butter was bitter"
words = s.split()
l = []
k = []
unique_words = sorted(set(words))
for word in unique_words:
k.append(word)
l.append(words.count(word))
z = zip(k,l)
print z
z = sorted(z, key=lambda x: x[1] , reverse=True)
print z
sorted() does not sort in place and it has a built-in reverse option that you omitted.
To count the words in the string, you can simply use Counter from collections. Then sort it in the descending order of counts.
Your code can be shortened to
from collections import Counter
s = "betty bought a bit of butter but the butter was bitter"
c = Counter(i for i in s.split())
print sorted(c.items(),key=lambda x:x[1],reverse=True)
zlist of word:count tuples, followed by the last three lines.