[ad_1] To combine two items of two lists as string you need to iterate through both lists at the same time and concatenate the two items as strings. list1 = [1, 2, 3, 4] list2 = [‘a’, ‘b’, ‘c’, ‘d’] list3 = [str(x) + str(y) for x, y in zip(list1, list2)] print(list3) the output will … Read more
[ad_1] Here’s a modification on @schwobasggl’s answer, using f-strings that works in Python 3.6+ def up_and_down(n): # 1,2, …, n, …, 2, 1 return (*range(1, n), *range(n, 0, -1)) def diamond(n, pad=’#’, fill=”*”): w = (2 * n) – 1 for i in up_and_down(n): # you can also use the below, but it’s slightly less … Read more
[ad_1] Use zip and slices >>> for i,j in zip(en[::2],en[1::2]): … print(“{}{}”.format(i,j)) … 12 34 56 As Steven Rumbalski mentions in a comment you can also do >>> it = iter(en) >>> for i,j in zip(it, it): … print i,j … 1 2 3 4 5 6 it here is an iterator over the list. … Read more
[ad_1] When describing a list, always consider using DCGs. In your case, you can very easily obtain what you want with a few simple modifications to your code: show_result(Squares,MaxRow,MaxCol, List) :- phrase(show_result(Squares,MaxRow,MaxCol,1), List). show_result(_,MaxRow,_,Row) –> { Row > MaxRow }, !. show_result(Squares,MaxRow,MaxCol,Row) –> { phrase(show_result(Squares,MaxRow,MaxCol,Row,1), Line) } , [Line], { Row1 is Row+1 }, show_result(Squares,MaxRow,MaxCol,Row1). … Read more
[ad_1] Just use zip to get tuples of values at corresponding indices in the two lists, and then cast each tuple to a list. So: [list(t) for t in zip(list1, list2)] is all you need to do. Demo: >>> list1 = [“X”, “Y”, “Z”] >>> list2 = [1, 2, 3] >>> list3 = [list(t) for … Read more
[ad_1] items = [1, 2, 3, 4, 5, 6, 7, 8, 9] for item in items: if 5 <= item <= 10: print(item) You might find Python for Beginners useful. 0 [ad_2] solved Search list for number in range
[ad_1] from collections import defaultdict d = defaultdict(list) for i,j in enumerate(x): d[j].append(i) Using defaultdict with the default being empty list and enumerate to provide the index [ad_2] solved Python: build dictionary from list
[ad_1] I’m trying to give an answer that will help you understanding the problem and work on the solution. You have a string with capital letters and at some point there is a small letter. You want the string to be split at the position before the first small letter. You can iterate through the … Read more
[ad_1] Simplest will be bubble sort. item* sort(item *start){ item *node1,*node2; int temp; for(node1 = start; node1!=NULL;node1=node1->next){ for(node2 = start; node2!=NULL;node2=node2->next){ if(node2->draw_number > node1->draw_number){ temp = node1->draw_number; node1->draw_number = node2->draw_number; node2->draw_number = temp; } } } return start; } 1 [ad_2] solved Assorting linked list in c [duplicate]
[ad_1] You can use pattern matching to determine which type of Array you’re dealing with. def arrTest[A](a :Array[A]) :Boolean = a match { case sa :Array[String] => sa.length < 2 || sa.sliding(2).forall(x => x(0) < x(1)) case ia :Array[Int] => ia.length > 2 && ia(0) == 1 && ia(1) == 1 && ia.sliding(3).forall(x => x(0) … Read more
[ad_1] I think that is what you need: s=”Here Comes The Sun And I Say It Is Alright”.split() for i in range(2, len(s), 3): s[i] = s[i – 1] print(‘ ‘.join(s)) # ‘Here Comes Comes Sun And And Say It It Alright’ 8 [ad_2] solved Removing words from a string- Python 2.7
[ad_1] You cannot achieve this if you want to stay with your custom type. Reason is that if you define a type with some properties then when initializing that type the properties are there, whether you assign to them a value or not. However using anonymous types and dynamic you can: var result = data.Select(item … Read more
[ad_1] The Main approach for checking elements of a list in a string is : s=””‘<a href=”https://ertfwetwer” target=”_blank”>[Nerve Center]</a>”’ my_list=[‘href’,’a’] def checker(mylist, my_string) new = list() for i in mylist: if i in my_string: # if elements is in string (you can check only special elements ) print i ,’is in string’ new.append(i) #storing result … Read more
[ad_1] They only reason to use method two is because you want to create a copy of arr2. Changes to the copy would not affect the original. Python 3.6.8 (default, Feb 14 2019, 22:09:48) [GCC 7.4.0] on cygwin Type “help”, “copyright”, “credits” or “license” for more information. >>> arr1 = [] >>> arr2 = [1,2,3,4,5] … Read more
[ad_1] This is the way to go: x = [‘2’, ‘3/4’, ‘+’, ‘4’, ‘3/5’] g = [] for i in x: try: g.append(int(i)) except ValueError: pass print(g) # [2, 4] What you tried is not a try-except block. It is an if statement which failed because ‘4’ is not an int instance (4 is, notice … Read more