After a couple of tryouts I just achieved to solve it: tuple_1 = (‘asdf@’, ‘asdf’) tuple_2 = () for elem in [item for item in tuple_1 if ‘@’ in item]: tuple_2 = tuple_2 + (elem,) print tuple_2 I hope this is what you need! I’m voting you up, since it was a little bit of … 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
As suggested in comments, you could use itertools.groupby to subdivide the tuple into “not a separator” and “separator” groups and then pick those groups that are not separators. This works for arbitrary numbers of segments, but can also be unpacked into two variables if you know to have exactly two groups. >>> a = (‘a’, … Read more
Are you looking for this? var tupleList = new List<Tuple<string, string>>(); for (int i = 0; i < data.Count; i++) { for (int j = i + 1; j < data.Count; j++) { tupleList.Add(new Tuple<string, string>(data[i], data[j])); } } 1 solved Combine elements of a list [closed]
The problem is that when you sort the GPAs, they’re no longer associated with the name ordering – you can simply skip doing so and the list indicies will stay correct or bring them into some collection (the instructions actually specify a tuple) # starting values scores_GPA = [3.85, 4.89, 2.89, 1.02, 3.8, 3.7, 2.9, … Read more
t = list(t) list(t) returns a new list while keeping t as a tuple. Think about a case you want to have both tuple and list. t2 = list(t) t is still a tuple while t2 is now a list. solved Permanently converting tuples to lists in Python [closed]
Hoo boy, this was a big XY problem. You want to sort by the third character of the second element of each tuple. Use the sorted built-in with its kwarg key. lst = [(‘abc’, ‘bcd’), (‘abc’, ‘zza’)] result = sorted(lst, key=lambda t: t[1][2]) key here accepts a function that is given each element in turn, … Read more
If time ends with – b then you could use normal str.split(“- b”) to get parts. And you would have to run it in loop results = [] for line in all_lines: parts = line.split(‘ – b’) results.append( parts ) or if you want to modify time or data results = [] for line in … Read more
you can use basic for loop and if statement to check values: List_1 = [((1.1, 2, 3), (1.1, 2, 3, 4), (3, 4, 5), 5, 6, 7)] List_2 = [((1.1, 2, 3), (1.1, 2, 3, 4), (3, 4.4, 5), 5, 6, 7)] List_1 = list(List_1[0]) # [(1.1, 2, 3), (1.1, 2, 3, 4), (3, 4, … Read more
Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed] solved Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed]
The .0 in somePoint.0 is accessing the first element (at index 0) of the tuple somePoint. .1 is accessing the second element (at index 1). As others have pointed out, this is covered in the first section of the language guide, “The Basics”. 3 solved What does the .0 and .1 mean Swift 3.0.1
This code works correctly on clang and g++ (using C++11): http://coliru.stacked-crooked.com/a/c8071ab447e10a31 Produces a std::tuple with Max_N elements of the same type, and fills it up with the first values of the given list. If less elements in list than Max_N, fills the last elements with sentinel values. In practice, using std::array<C, N> may be more … Read more
Since your tuples are just repeating the same number, just use a range loop and create the tuple from scratch: for i in range(256): rgb = (i, i, i) There is no need to add up here. Alternatively, calculate i from the frame. 30 frames in a second? Divide 256 by 30 to get a … Read more
First, you should be more specific of what you expect. If you want your list to be the sum of the numbers in the tuples, here’s what you can do: listnum = [sum(x) for x in listnum] 1 solved How can I add tuples in a list? [closed]
Here’s a quick function to do this looping through the tuples and comparing each tuple to the label of the previous tuple and concatenating them if the labels match. def parse_tuples(x): prev_tuple = list(x[0]) parsed = [] for i in x[1:]: if i[1] == prev_tuple[1]: prev_tuple[0] += i[0] else: parsed.append(tuple(prev_tuple)) prev_tuple = list(i) parsed.append(tuple(prev_tuple)) return … Read more