WebPython 3.3, a dictionary with key-value pairs in this form. d = {'T1': ['eggs', 'bacon', 'sausage']} The values are lists of variable length, and I need to iterate over the list items. This works: count = 0 for l in d.values (): for i in l: count += 1 But it's ugly. There must be a more Pythonic way, but I can't seem to find it. len (d.values ()) WebMay 3, 2024 · The keys are words the value is the number of times those words occur. countDict = {'house': 2, 'who': 41, 'joey': 409, 'boy': 2, 'girl':2} I'd like to find out how many elements occur with a value of more than 1, with a value of more than 20 and with a value of more than 50. I found this code
def load_dict (dict_name): if os.path.isfile (os.path.join (
Web注意: C:\Python 是Python的安装目录。也可以通过以下方式设置：右键点击"计算机"，然后点击"属性"然后点击"高级系统设置"选择"系统变量"窗口下面的"Path",双击即可！然后在"Path"行，添加python安装路径即可(我的D:\Python32)，所以在后面，添加该路径即 可。 WebMay 24, 2024 · For this you could write the following function that would work for data in the structure you provided (a list of dicts): def count_key (key,dict_list): keys_list = [] for item in dict_list: keys_list += item.keys () return keys_list.count (key) Then, you could invoke the function as follows: see you on down the road
Python Count number of items in a dictionary value that is a list
WebJun 21, 2015 · 3 Answers Sorted by: 23 Use collections.Counter and its most_common method: from collections import Counter def predominant_sign (data): signs = Counter (k ['sign'] for k in data if k.get ('sign')) for sign, count in signs.most_common (): print (sign, count) Share Improve this answer Follow answered Jun 21, 2015 at 10:17 vaultah 43.2k … WebTo count unique values per key, exactly, you'd have to collect those values into sets first: values_per_key = {} for d in iterable_of_dicts: for k, v in d.items (): values_per_key.setdefault (k, set ()).add (v) counts = {k: len (v) for k, v in values_per_key.items ()} which for your input, produces: WebThe keys () method returns a view object. The view object contains the keys of the dictionary, as a list. The view object will reflect any changes done to the dictionary, see … see you now podcast apple