Question

我正在2个牌组中搜索常用术语。我想搜索卡片组[v] ... 1到卡片组[b] ... 1,2,3,4,5,6,7,8,9,10,11,12,13。然后，我想将[v]递增以搜索卡片组[v] ... 2到卡片组[b] ... 1,2,3,4,5,6,7,8,9,10,11,12， 13 等等等我可以获得第一部分，但不知道如何增加[v]并将[b]重置为1。任何帮助，将不胜感激。有13个字词

b = 1
v = 1
while b <= 13:
    print(b, deck[b])
    print("Intersection between", "v = ",v, "and", "b = ",b, deck[v].intersection(deck[b]))
    if len(deck[v].intersection(deck[b])) == 1:
        print("1 match")
        print("v = ", v)
        print("b = ", b)
    b = b + 1

Answer 1

为什么不只使用2个循环？

for b in range(1,14):
 for v in range(1,14):
   # Your code
   print(b, deck[b])
   print("Intersection between", "v = ",v, "and", "b = ",b, deck[v].intersection(deck[b]))
   if len(deck[v].intersection(deck[b])) == 1:
      print("1 match")
      print("v = ", v)
      print("b = ", b)

Answer 2

您只需要将当前代码插入新循环并在该循环中递增v：

b = 1
v = 1
while v <= 13:
    while b <= 13:
        print(b, deck[b])
        print("Intersection between", "v = ",v, "and", "b = ",b, deck[v].intersection(deck[b]))
        if len(deck[v].intersection(deck[b])) == 1:
            print("1 match")
            print("v = ", v)
            print("b = ", b)
        b = b + 1
    v = v + 1

如何增加一个值并重设另一个值。蟒蛇

2 个答案: