CODEWITHSUNDEEP
pythonpython-2.7dictionarylist-comprehension
Alby11
Alby11

Reputation: 645

python and list/dict comprehension

I'm learning some list/dict comprehension and I'm stuck!!!

I really don't understand the following...

I have this program:

def histogram_for(word):
    adict = dict()
    for c in word:
        adict[c] = adict.get(c, 0) + 1
    print adict
    return adict

def histogram_dc(word):
    adict = dict()
    adict = {c: (adict.get(c, 0) + 1) for c in word}
    print adict
    return adict

def histogram_lc(word):
    adict = dict()
    adict = dict([(c, (adict.get(c, 0) + 1)) for c in word])
    print adict
    return adict


word = "Football"
histogram_for(word.lower())
histogram_dc(word.lower())
histogram_lc(word.lower())

And i get these results:

{'a': 1, 'b': 1, 'f': 1, 'l': 2, 'o': 2, 't': 1}
{'a': 1, 'b': 1, 'f': 1, 'l': 1, 'o': 1, 't': 1}
{'a': 1, 'b': 1, 'f': 1, 'l': 1, 'o': 1, 't': 1}

Why the only working one is the "for" method?

Upvotes: 2

Views: 174

Answers (2)

Sean Vieira
Sean Vieira

Reputation: 159865

Quite simply because while the processing is happening in _dc and _lc adict is empty, while in _for it's being updated on each turn of the for loop. A comprehension can be de-sugared into a for loop of its own:

adict = dict()
adict = {c: (adict.get(c, 0) + 1) for c in word}

becomes:

adict = dict()
# Handwaving for what Python actually does
$newdict = {}
for c in word:
    $newdict[c] = adict.get(c, 0) + 1
adict = $newdict

Use collections.Counter (or the for-loop version) if you need to keep track of a set of keys and counts of occurrences.

Upvotes: 3

cromod
cromod

Reputation: 1799

As Sean Vieira said, the class collections.Counter and its method most_common are the best solution to your need.

But, if you really want to keep list/dict comprehension, I suggest using set and count : 

def histogram_dc(word):
  adict = dict()
  adict = {c: word.count(c) for c in set(word)}
  print adict
  return adict

def histogram_lc(word):
  adict = dict()
  adict = dict([(c, word.count(c)) for c in set(word)])
  print adict
  return adict

Upvotes: 1

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