蟒紋消除多次打印輸出

Question

代碼：

x = ['1', '2', '3'] 
y = ['a', 'b', 'c'] 
rangeend = len(x) 
for i in range(0, rangeend): 
    with open("file20.txt") as f: 
     for line in f: 
      count = 0 
      line = line.strip() 
      z = line.split(" ") 
      if z[0] == x[i] : 
       count = 1 
       a = z[0], z[1] 
       b = x[i],y[i] 
       if a == b: 
        print "ok" 
        break 
       else: 
        print "Failed" 
        break 
      if count != 1: 
       print "{} not found".format(x[i]) 

file20.txt：

1 a 
2 b 
5 c 

x列表中有3但file20.txt沒有在開始該行的3（第一串）

我正在嘗試的是我要打印的文件是3 not found，並且打印只能完成一次在最後。

注意：不僅元素3，任何元素如果x列表有但file20.txt文件行不在開始（第一個字符串）。我想要打印沒有找到。

下面是代碼輸出和2實際上是存在的，但它是印刷2 not found（其實這不打印2 not found）和3是不存在的file20.txt但打印3 not found是正確的，但問題是，它是印刷3倍。我只是想要它打印3 not found，只有一次。

0 
('1', 'a') ('1', 'a') 
ok 
1 
2 not found 
('2', 'b') ('2', 'b') 
ok 
2 
3 not found 
3 not found 
3 not found 

Answer 1

移動if count != 1塊內for循環外看起來像工作。你有沒有試過！

x = ['1', '2', '3', '4'] 
y = ['a','b','c','d'] 
rangeend = len(x) 
for i in range(0, rangeend): 
    with open("bob.txt") as f: 
     for line in f: 
      count = 0 
      line = line.strip() 
      z = line.split(" ") 
      if z[0] == x[i] : 
       count = 1 
       a = z[0], z[1] 
       b = x[i], y[i] 
       if a == b: 
        print "ok" 
        break 
       else: 
        print "Failed" 
        break 
     if count != 1: 
      print "{} not found".format(x[i])