Python UserDict.DictMixin.iteritems() Examples

def elements(self):
            '''Iterator over elements.

            It repeats each element as many times as its count.

            >>> c = Counter('ABCABC')
            >>> sorted(c.elements())
            ['A', 'A', 'B', 'B', 'C', 'C']

            If an element's count has been set to zero or is a negative
            number, elements() will ignore it.
            '''
            for elem, count in self.iteritems():
                for _ in repeat(None, count):
                    yield elem

        # Override dict methods where the meaning changes for Counter objects. 

def elements(self):
            '''Iterator over elements.

            It repeats each element as many times as its count.

            >>> c = Counter('ABCABC')
            >>> sorted(c.elements())
            ['A', 'A', 'B', 'B', 'C', 'C']

            If an element's count has been set to zero or is a negative
            number, elements() will ignore it.
            '''
            for elem, count in self.iteritems():
                for _ in repeat(None, count):
                    yield elem

        # Override dict methods where the meaning changes for Counter objects. 

def iteritems(self, raw=False):
        """ Iterate self items. """

        for key in self:
            yield key, self.__getitem__(key, raw=raw) 

def most_common(self, n=None):
            '''List the n most common elements and their counts.

            The list goes from the most common to the least.  If n is
            None, then list all element counts.

            >>> Counter('abracadabra').most_common(3)
            [('a', 5), ('r', 2), ('b', 2)]
            '''
            if n is None:
                return sorted(self.iteritems(), key=itemgetter(1),
                              reverse=True)
            return nlargest(n, self.iteritems(), key=itemgetter(1)) 

def update(self, iterable=None, **kwds):
            '''Like dict.update() but add counts instead of replacing them.

            Source can be an iterable, a dictionary, or another Counter
            instance.

            >>> c = Counter('which')
            >>> c.update('witch')      # add elements from another iterable
            >>> d = Counter('watch')
            >>> c.update(d)            # add elements from another counter
            >>> c['h']                 # four 'h' in which, witch, and watch
            4
            '''
            if iterable is not None:
                if hasattr(iterable, 'iteritems'):
                    if self:
                        self_get = self.get
                        for elem, count in iterable.iteritems():
                            self[elem] = self_get(elem, 0) + count
                    else:
                        # fast path when counter is empty
                        dict.update(self, iterable)
                else:
                    self_get = self.get
                    for elem in iterable:
                        self[elem] = self_get(elem, 0) + 1
            if kwds:
                self.update(kwds) 

def most_common(self, n=None):
            '''List the n most common elements and their counts.

            The list goes from the most common to the least.  If n is
            None, then list all element counts.

            >>> Counter('abracadabra').most_common(3)
            [('a', 5), ('r', 2), ('b', 2)]
            '''
            if n is None:
                return sorted(self.iteritems(), key=itemgetter(1),
                              reverse=True)
            return nlargest(n, self.iteritems(), key=itemgetter(1)) 

def update(self, iterable=None, **kwds):
            '''Like dict.update() but add counts instead of replacing them.

            Source can be an iterable, a dictionary, or another Counter
            instance.

            >>> c = Counter('which')
            >>> c.update('witch')      # add elements from another iterable
            >>> d = Counter('watch')
            >>> c.update(d)            # add elements from another counter
            >>> c['h']                 # four 'h' in which, witch, and watch
            4
            '''
            if iterable is not None:
                if hasattr(iterable, 'iteritems'):
                    if self:
                        self_get = self.get
                        for elem, count in iterable.iteritems():
                            self[elem] = self_get(elem, 0) + count
                    else:
                        # fast path when counter is empty
                        dict.update(self, iterable)
                else:
                    self_get = self.get
                    for elem in iterable:
                        self[elem] = self_get(elem, 0) + 1
            if kwds:
                self.update(kwds) 

def kvquery(root, **kwargs):
    '''kvquery provides a convinient way of finding widgets in an
    application that uses the kv style language.

    example:
        lets say you have a .kv file with the following Rule:
        <MovieWidget>:
            BoxLayout:
                Video:
                    kvid: 'video'
                Label:
                    text: root.movie_title
                Label:
                    text: root.movie_description

        in your python code, you may want to get the reference to
        Video widget nested inside the widget you have a handle to.

        # video will be the first node that jas a 'kvid' property == 'video'
        video = kvquery(movie, kvid='video').next()


        #lets get all teh labels in a list
        labels = list(kvquery(movie, __class__=Label))


    :Parameters:
    `root`: root of the tree to queried
        this node and all decendants will be iterated by the
        returned generator.

    `**kwargs`: **kwargs, key/value pairs
        The keys corrosponf to porperty names, and values to the
        property values of the widget nodes being queried.  If a node
        has at least one attr such that (gettattr(node, key) == value)
        is true; it will be included in the iteration.
    '''

    def _query(w):
        '''iternal query function / predicate for tree query
        '''
        for k, v in kwargs.iteritems():
            if (v == getattr(w, k, None)):
                return True


    return filter_tree(root, _query)