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cheat codes, hints, tips, tricks and other secrets to make your playing experience the best! (Hacking) 5T DMB Subscriber Patch 3.0.0 C.Q: Python:



 

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Battlefield 1942: The download link. With the time that you have allowed me. SolucionarioCalorYTermodinamicaZemansky Cheatbook.com cheats, codes, cheat codes, hints, tips, tricks and other secrets to make your playing experience the best! (Hacking) 5T DMB Subscriber Patch 3.0.0 C.Q: Python: exception caught when splitting two lists of lists in different places While I was making a program, I got to this part, which fails, by making the interpreter throw an exception: >>> o3 = [] >>> for i in range(3): ... a = list(range(i+1, 3)) ... o3.append(a) ... >>> o3 [list(range(1, 3)), list(range(2, 3)), list(range(3, 3))] >>> list(o3) [None, None, None] So, I guess I do something wrong, and as I've already seen, is to do with the append() function. However, I can't seem to find the error, as the program doesn't make sense in any way. A: The problem is that you create three lists in o3 that each contain a reference to the same list object. You can fix it by creating a copy: >>> o3 = [] >>> for i in range(3): ... a = list(range(i+1, 3)) ... o3.append(a) ... >>> list(o3) [list(range(1, 3)), list(range(2, 3)), list(range(3, 3))] I think you can also do: >>> from copy import deepcopy ... o3 = [] >>> for i in range(3): ... a = deepcopy(list(range(i+1, 3))) ... o3.append(a) ... >>> list(o3) [list(range(1, 3)), list(range(2, 3)), list(range(3, 3))] Q: Do metacharacters mean in a regex I have seen this




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