Not sure if I got your goal correctly but based on the examples you gave the following approach might work. CC = {} for step in range(len(Nodes[‘time’])): for key in CoordComboSort.keys(): CC[Nodes[‘time’][step]] = {key : CoordComboSort[key][step] for key in CoordComboSort.keys()} For your input, the output will be like this: {‘A’: {0: (1, 4, 5), 0.001: … Read more
You can use this list of words https://github.com/dwyl/english-words var input = “hello123fdwelcome”; var fs = require(‘fs’); fs.readFile(“words.txt”, function(words) { var words = words.toString().split(‘\n’).filter(function(word) { return word.length >= 4; }); var output = []; words.forEach(word) { if (input.match(word)) { output.push(word); } }); console.log(output); }); solved Javascript: find english word in string [closed]
Not beautiful solution, but working one values = [{‘id’: 1, ‘x’: 2}, {‘id’: 1, ‘y’: 4}, {‘id’: 1, ‘z’: 6}, {‘id’: 2, ‘j’: 5}, {‘id’: 2, ‘k’: 10}, {‘id’: 3, ‘w’: 1}, {‘id’: 3, ‘x’: 3}, {‘id’: 3, ‘y’: 5}, {‘id’: 3, ‘z’: 7}] # get all unique possible keys unique_keys = set((x[‘id’] for x … Read more
This data structure is very difficult to deal with. I would consider if this is the best way to store this data. Maybe an array of structs that represent your data? The confusing part is your variable is named array, but it is really declared as a Dictionary literal. So to access the item with … Read more
You seem to be approaching this as if it is C. That is understandable, but overcomplicating your Python. Let’s consider just a few things. Overwriting an input variable immediately: def buildIndex(m, d): d = {} In Python we don’t need to declare variables or make room for them. So this is much more clearly expressed … Read more
There are a few different components to this problem. First, how do you split the receipt into each individual company. Next you need to be able to parse the company’s ID. Finally you need to be able to parse the quantity, cost, and total cost from a line item. Splitting Receipts Your method of splitting … Read more
You want something like the following: d = {x[:2]:’item’+str(i+1) for i, x in enumerate(my_list)} I’m going to break down how this works so it’s less “python voodoo” in the future: Firstly, we want the first two values from each tuple in the list. This is done by list slicing: x[:2]. This says “give me the … Read more
Accessing values in a dictionary requires brackets [], not parentheses () (see this answer. From the code you’ve given us, it looks like grads(“db” + str(l+1)) needs to be changed to grads[“db” + str(l+1)] to fix your error. A few notes on style: The character lowercase ‘l‘ shouldn’t be used as a variable name in … Read more
Maybe use zip: for a,b,c in zip(scores,boxes,class_list): if a >= 0.98: data[k+1] = {“score”:a,”box”:b, “class”: c } ; k=k+1; print(“Final_result”,data) Output: Final_result {2: {‘score’: 0.98, ‘box’: [0.1, 0.2, 0.3, 0.4], ‘class’: 1}} Edit: for a,b,c in zip(scores,boxes,class_list): if a >= 0.98: data[k+1] = {“score”:a,”box”:b, “class”: int(c) } ; k=k+1; print(“Final_result”,data) 9 solved Appeding different list … Read more
Assuming that you want to name your keys as key1, key2.. and so on when a key is not found: l = [[‘a’,1], [‘b’,2], [3], [‘d’,4]] d = {} i = 1 for x in l: try: d[x[0]] = x[1] except IndexError: d[‘key’+str(i)] = x[0] i += 1 print(d) Output: {‘a’: 1, ‘b’: 2, ‘key1’: … Read more
I believe it should just be a couple of small typos. First, your second line looks like it should be indented, and where you have links[“property”], it looks like you would want link[“property”]. for link in links: payload[f’Links[{link[“property”]}][url]’] = link[‘url’] The reason it is giving you that error is that you are trying to get … Read more
Keeping it simple, you could do something like this: reading = 0 notReading = 0 for book in bookLogger: if book[‘Process’] == ‘Reading’: reading += 1 elif book[‘Process’] == ‘Not Reading’: notReading += 1 print(f’Reading: {reading}’) print(f’Not Reading: {notReading}’) Alternatively, you could also use python’s list comprehension: reading = sum(1 for book in bookLogger if … Read more
If you mean that you have several fields like 98 but they all contain a title, you could do this: titles = list() for k in my_dict[“countries”].keys(): if my_dict[“countries”][k].has_key(“title”): titles.append(my_dict[“countries”][k][“title”]) or, as suggested in the comments try: titles = [item[‘title’] for item in dictName[‘countries’]] except KeyError: print(“no countries/title”) solved How to get specific value in … Read more
def map(func, values): rest = values[1:] if rest: yield from map(func, rest) else: yield func(values[0]) It’s a generator so you can iterate over it: for result in map(int, ‘1234’): print(result + 10) Gives: 11 12 13 14 4 solved how to implement map and reduce function from scratch as recursive function in python? [closed]
If you have a IEnumberable of KeyValuePairs then you could do something like this: public bool AreAllSame(IEnumberable<KeyValuePair<Int64, MyObject>> list) { return list.Select(kv => kv.Value).Distinct().Count == 1; } Not sure whether it’s really optimized but, it’s shorter! :-] Of course, whatever you’re comparing will need to be comparable. 2 solved Optimizing foreach loop that compares multiple … Read more