Function count

pattern/text/en/inflect_quantify.py:302–330

Returns an approximation of the entire set. Identical words are grouped and counted and then quantified with an approximation.

(*args, **kwargs)

# count([word1, word2, ...], plural={})
# counr({word1:0, word2:0, ...}, plural={})
def count(*args, **kwargs):
    """ Returns an approximation of the entire set.
        Identical words are grouped and counted and then quantified with an approximation.
    """
    if len(args) == 2 and isinstance(args[0], basestring):
        return approximate(args[0], args[1], kwargs.get("plural", {}))
    if len(args) == 1 and isinstance(args[0], basestring) and "amount" in kwargs:
        return approximate(args[0], kwargs["amount"], kwargs.get("plural", {}))
    if len(args) == 1 and isinstance(args[0], dict):
        count = args[0]
    if len(args) == 1 and isinstance(args[0], (list, tuple)):
        # Keep a count of each item in the list.
        count = {}
        for word in args[0]:
            try:
                count.setdefault(word, 0)
                count[word] += 1
            except:
                raise TypeError, "can't count %s, only str and unicode" % word.__class__.__name__
    # Create an iterator of (count, item) tuples, sorted highest-first.
    s = [(count[word], word) for word in count]
    s = max([n for (n,w) in s]) > 1 and reversed(sorted(s)) or s
    # Concatenate approximate quantities of each item,
    # starting with the one that has the highest occurence.
    phrase = []
    for i, (n, word) in enumerate(s):
        phrase.append(approximate(word, n, kwargs.get("plural", {})))
        phrase.append(i==len(count)-2 and " and " or ", ")
    return "".join(phrase[:-1])

quantify = count