Question

代码：

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列表包含3但file20.txt在行的开头没有3（第一个字符串）

我想要打印的是文件中的3 not found，并且最后只需要打印一次。

注意：不仅元素3，x列表中的任何元素，但file20.txt文件行在开头没有（第一个字符串）。我想要找不到印刷品。

下面是代码输出，2实际存在，但它正在打印2 not found（实际上它不会打印2 not found）3不在file20.txt中1}}但打印3 not found是正确的，但问题是打印3次。我只想让它打印3 not found，只打一次。

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

将if count != 1块移到内部for循环之外看起来像是在工作。你试过了吗？

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])

python print消除了更多次打印到输出

1 个答案: