[ad_1] l is pointing to the head of a linked list. So l is assigned NULL to start with, i.e. empty linked list. In the first iteration of the for-loop a node is created and assigned the value “06”. l is set to point to the newly created node. So you have: l —> (“06”, … Read more
[ad_1] Use regex a=”HOW TO COOK RICE – Chapter 1: 1 rinse water once 2 add one and a half cup of water for every cup of rice 3 cover the pot and simmer” idx=[i.start() for i in re.finditer(‘(\d{1,} \w{1,})’, a)] if idx: print(a[:idx[0]]) else: print(a) for i,index in enumerate(idx): if not i==len(idx)-1: print(a[index:idx[i+1]]) else: … Read more
[ad_1] you can use the Sort overload accepting a Comparison. This should work: public static int MyCompare(double x, double y) { if (x >= 0.0 == y>=0.0) // same sign, compare by absolute value return Math.Abs(x).CompareTo(Math.Abs(y)); if (x < 0.0) return 1; return -1; } usage: var list = new List<double>(); // fill your list … Read more
[ad_1] dictA = {“f1” : [“m”], “f2” : [“p”,”y”,”o”,”a”,”s”,”d”,”f”,”g”], “f3” : [“w”,”t”], “f5” : [“z”,”x”], “f6” : [“c”,”v”]} result = [] limit_size = 3 values_list = [] for values in dictA.itervalues(): values_list.append(len(values)) for i in range(0,max(values_list)): keys = list(dictA.keys()) count = 0 while count < len(keys): try: result.append(dictA[keys[count]][i]) except IndexError: dictA.pop(keys[count]) count = count + … Read more
[ad_1] Here is a very compact implementation: let to_str_list = List.map (fun cs -> String.concat “” (List.map (String.make 1) cs)) It looks like this when your call it: # to_str_list [[‘h’;’e’;’l’;’l’;’o’];[‘w’;’o’;’r’;’l’;’d’]];; – : string list = [“hello”; “world”] Update If you want to suppress empty strings at the outer level you can do something like … Read more
[ad_1] Hints I would follow this steps: Create a list where I will store my partial strings Start iterating the string Store the initial position and the current character Keep iterating until the character is different Store in the list the partial string from the initial position you stored until 1 less than the current … Read more
[ad_1] Yes, there is the split function, however before calling it on your string, you must get rid of the [ and ], or else instead of [‘1’, ‘2’, ‘3’, ‘4’], you will get [‘[1’, ‘2’, ‘3’, ‘4]’], so instead of s.split(), we do s[1:-1].split(), also this means your list is strings instead of ints … Read more
[ad_1] I think there are a couple of misunderstandings here. First, most types in OCaml are immutable. Unless you use mutable variables you can’t “remove it from the list”, you can only return a version of the list that doesn’t have that first item. If you want to return both things you can achieve that … Read more
[ad_1] >>> l = [0.01, 0.1, 0.4, 0.034, 0.6, 0.7, 0.9, 1] >>> [0.4 if (0. < f < 0.5) else 0.7 for f in l] [0.4, 0.4, 0.4, 0.4, 0.7, 0.7, 0.7, 0.7] [ad_2] solved Change the numbers of a list to another number python
[ad_1] Here is a method to do what you are trying to achieve. unsorted = [[1, 0], [2, 0], [3, 0], [4, 2], [5, 2], [6, 5], [7, 6], [8,0]] sortList = [] sortDict = {} for x in unsorted: if x[1] != 0: if x[1] in sortDict: sortDict[x[1]].append(x[0]) else: sortDict[x[1]] = [x[0]] for x … Read more
[ad_1] You can use ast.literal_eval to safely evaluate strings containing Python literals. from ast import literal_eval a=”[“Hello”, “World!”, 2]” b = literal_eval(a) # [“Hello”, “World!”, 2] Note that the string can only be compromised of: strings, bytes, numbers, tuples, lists, dicts, sets, booleans, and None (taken from the documentation here) [ad_2] solved how to convert … Read more
[ad_1] >>> from itertools import groupby >>> from operator import itemgetter >>> list1 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> list2 [0, 0, 3, 4, 5, 0, 0, 8, 9, 0] >>> [[e[0] for e in v] for k,v in groupby(((a ,b, a==b) for a,b in zip(list1, list2)), itemgetter(2)) if k] … Read more
[ad_1] You are trying to change List<String> to List<tcmb> and problem lies here public static List<tcmb> getList(Activity a){ // your code // List<String> itemList = new ArrayList<String>(); itemList.addAll(Arrays.asList(itemWords)); dovizList = (List)itemsList; Log.d(TAG, “getValuestcmb: ” + dovizList.size()); return dovizList; Also, I don’t understand, what exactly you are trying to achieve here. List<String> itemsList = new ArrayList<String>(); … Read more
[ad_1] Now that I understand that the list A itself is not going to change, but you expressed the idea of change meaning that the list has partitions where there are adjacent elements that are different, then a different solution comes to mind: Given your original data: A = [‘1′,’1′,’1′,’3′,’3′,’4’] B = [‘5′,’9′,’3′,’4′,’8′,’1’] We can … Read more
[ad_1] 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 … Read more