Use Counter >>> from collections import Counter >>> c=Counter() >>> for d in l: … c.update(d) … >>> dict(c) {‘a’: 30, ‘b’: 15} 1 solved python : Get the sum of values of a key in list of dictionaries [duplicate]
import itertools as IT items = [‘Thu Apr 04’, ‘ Weigh In’, ‘Sat Apr 06’, ‘ Collect NIC’, ‘ Finish PTI Video’, ‘Wed Apr 10’, ‘ Serum uric acid test’, ‘Sat Apr 13’, ‘ 1:00pm’, ‘Get flag from dhariwal’, ‘Sun Apr 14’, ‘ Louis CK Oh My God’, ‘ 4:00pm’, ‘UPS Guy’] date_word = (‘Mon’, … Read more
You can use List#grouped to create a non-overlapping window: list.grouped(2).map(l => (l(0), l(1))).toMap 7 solved How to convert a list of string to map in Scala?
You need to actually initialize syncModelRecord. Try this: let syncModelRecord = SyncModelRecord(date: String(Int64(syncTimestamp!)), shakeState: 0) dummySyncModelRecordArray.append(syncModelRecord!) Another tip maybe worth exploring, you can directly decode a date and specify the decoder’s strategy for it (in your case secondsSince1970) solved How to make an array using dummy values from Struct Modal class in Swift iOS
function convert(data){ var myMap = {} data.forEach(el => myMap[el] = myMap[el] != undefined ? myMap[el] + 1 : 1); return Object.keys(myMap).map(k => {return {name: k, y: myMap[k]}}) } console.log(convert([‘A’,’B’,’B’,’B’,’C’,’C’,’A’,’B’])) 2 solved JavaScript array to dictionary conversation based on count
You can use the dict constructor d = dict(key1=val1, key2=val2) or __setitem__ method d = {} d[‘key1’] = val1 d[‘key2’] = val2 There are other, even more flexible ways, but those are the basics. solved Generate Dictionary in Python at run time [closed]
I assumed date is increasing order. # store values tot = {} # the last date date0 = Dict1[-1][‘date’] # easier to work from back, i found for line in Dict1[-1::-1]: date, name, qty = [line[x] for x in ‘date’, ‘name’, ‘qty’] # add the value to all subsequent days for d in range(date, date0+1): … Read more
from itertools import product my_dict = {‘a’:1, ‘chk’:{‘b’:2, ‘c’:3}, ‘e’:{‘chk’:{‘f’:5, ‘g’:6}} } def process(d): to_product = [] # [[(‘a’, 1)], [(‘b’, 2), (‘c’, 3)], …] for k, v in d.items(): if k == ‘chk’: to_product.append([(k2, v2) for d2 in process(v) for k2, v2 in d2.items()]) elif isinstance(v, dict): to_product.append([(k, d2) for d2 in process(v)]) else: … Read more
Assuming each name belongs to only one key, and parents are defined before their children, you need to do three tasks for each element: map its name to a yet incomplete child list map its ID to its name link it to its parent If parents and children are not ordered, i.e. a child can … Read more
outer_keys = filter(lambda x: my_key in x, my_dict.keys()) result = my_dict[next(outer_keys)][my_key] 1 solved Find keys in nested dictionary [closed]
Not sure what you are trying to achieve, but it looks like you get an array of objects from the backend and you want to turn it into an array of objects, but with different property names. That’s a simple Array.prototype.map(): const response = await fetch(url); const data = await response.json(); const questions = data.map((question) … Read more
First iterating over the outer values using key_level1 and val_level1, then iterating over inner values using key_level2, val_level2: for key_level1, val_level1 in r.items(): for key_level2, val_level2 in val_level1.items(): for val in val_level2.split(‘,’): # Example: # do something print(key_level1, key_level2, val) 4 solved Python iterate over multi value nested dictionary [closed]
Now that you’ve fixed the original problem, and fixed the next problem with doing the check backward, and renamed all of your variables, you have this: for match in dictionary: if any(match in value for value in sentences): print match And your problem with it is: The way I have the code written i can … Read more
The easiest solution would be to iterate through the intersection of the keys and add the corresponding balance and price as an entry in a new dictionary. The intersection isn’t necessary if you can guarantee that if a key exists in either of balance of price, it will exist in the other. As I cannot … Read more
If you model your JSON object with a struct and you unmarshal into a value of that, then you don’t need those ugly indices and type assertions, you can simply refer to struct fields. Note that you don’t have to be afraid of the response being complex, you only need to model the parts you … Read more