And yes , you can just pass with if/else and for loops : a = [[0,0,0,0,0], [0,1,1,0,0], [0,1,0,1,0], [0,0,1,0,0], [0,0,0,0,0]] to_replace = 1 replacement = 9 for i in range(len(a)): for x in range(len(a[i])): if a[i][x] == to_replace: for pos_x in range(i-1,i+2): for pos_y in range(x-1,x+2): try: if a[pos_x][pos_y] != to_replace: a[pos_x][pos_y] = replacement except … Read more
Try this: def check(list_): last = None for element in list_: if element == last: return True else: last = element return False solved Check if there is a number near the same number in a list – python [duplicate]
if ‘ddd’ or ‘eee’ in test is evaluated as: if (‘ddd’) or (‘eee’ in test): as a non-empty string is always True, so or operations short-circuits and returns True. >>> bool(‘ddd’) True To solve this you can use either: if ‘ddd’ in test or ‘eee’ in test: or any: if any(x in test for x … Read more
You want something like this: public static Pair FindClosest(IEnumerable<Pair> list, int value) { Pair closest = null; if (list != null && list.Count() > 0) { foreach (var pair in list) { if (value <= pair.Max) { closest = pair; break; } } if (closest == null) { closest = list.Last(); } } return closest; … Read more
Try using: d = {i[‘service’]: i[‘price’] for i in a} print([{‘service’: i, ‘price’: d.get(i)} for i in b]) Output: [{‘service’: ‘basketball’, ‘price’: None}, {‘service’: ‘yoga’, ‘price’: 30}, {‘service’: ‘soccer’, ‘price’: None}, {‘service’: ‘golf’, ‘price’: 40}] 0 solved Compare in Python two lists and create a new one that is combinated [closed]
It is not directly possible, because what must be on the left side of an assignment cannot be a function call. It can only be built from simple variables, data members, subscripts and commas, parentheses or square brackets. Best that can be done is to use a comprehension or a map on the right side: … Read more
You should try to do the complete code, but the base code is x = [[‘#’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’], [‘#’, ‘ ‘, ‘ ‘, ‘ ‘, ‘ ‘, ‘ ‘, ‘#’], [‘#’, ‘ ‘, ‘$’, ‘+’, ‘$’, ‘ ‘, ‘#’], [‘#’, ‘.’, ‘*’, ‘#’, ‘*’, ‘.’, ‘#’], [‘#’, ‘ ‘, ‘$’, ‘.’, … Read more
Something like this… List<String> result = Stream.of(“A”, “A”, “BB”, “C”, “BB”, “D”) .collect(Collectors.groupingBy( Function.identity(), Collectors.counting())) .entrySet() .stream() .filter(x -> x.getValue() == 1L) .map(Entry::getKey).collect(Collectors.toList()); System.out.println(result); // [C, D] 2 solved return the list with strings with single occurrence [closed]
There are many options, I describe some of them for you use Dictionary<int, string>Pros: very fast lookupCons: you can not have two string with same number, you don’t have a List var list2d = new Dictionary<int, string>(); list2d[1] = “hello”; list2d[2] = “world!”; foreach (var item in list2d) { Console.WriteLine(string.Format(“{0}: {1}”, item.Key, item.Value); } use … Read more
b = max(a) [1 if x==b else 0 for x in a] max(a) will return the maximum element in your list so you want to essentially replace all non-max values with 0 and all max values to 1. You can do this several ways but the above seems most to the point. Basically this is … Read more
from itertools import combinations, permutations perm=[] index = [9,7,6,5,2,10,8,4,3,1] perm.append(index) M = 3 slicer = [x for x in combinations(range(1, len(index)), M – 1)] slicer = [(0,) + x + (len(index),) for x in slicer] result = [tuple(p[s[i]:s[i + 1]] for i in range(len(s) – 1)) for s in slicer for p in perm] 1 … Read more
Here’s a straightforward solution with a simple loop: dict_x = {} for value in list_x: if isinstance(value, str): dict_x[value] = current_list = [] else: current_list.append(value) Basically, if the value is a string then a new empty list is added to the dict, and if it’s a list, it’s appended to the previous list. solved Convert … Read more
To just add a count you could keep a set of itertools.count objects in a defaultdict: from itertools import count from collections import defaultdict counters = defaultdict(lambda: count(1)) result = [(n, next(counters[n])) for n in inputlist] but this would add counts to all elements in your list: >>> from itertools import count >>> from collections … Read more
You forgot to call the conversion function for some parameter in your function. I tested and corrected your code, call it this way: is_verified = verified( convert(data[‘k_signer’]), data[‘d_pk’], convert(data[‘d_signer’]), data[‘sig’], convert(data[‘addr’]), convert(data[‘seed’]), convert(data[‘data’]) ) solved How can i fix concatenate tuple (not “list”) to tuple
The python function split() by default splits your text on spaces. You can add separator inside, for example: task.split(‘,’) which means this function is going to split your text in list by commas. See more examples on w3schools. solved how to print each value of list on next line in python [closed]