Question

抱歉在这里成为一个菜鸟，因为我是这个新手

我一直在尝试编写一个代码，该代码可以在行中找到某些单词，然后从第二个单词列表中匹配，以查看第二个单词是否存在于同一行中。仅当第一个单词不存在于一行时才打印行;这是片段和示例

keywords =("a","b","c","d")
keywords2 = ("e","f","g")
a = ("a  is a ", "a is not e","b is b", "b is not f")
for key2 in keywords2:              
    for key in keywords:                

        for line in a:

                if key2 not in line:
                    if key in line:

                        print(key+" Found in --> ")
                        print      (line)

所需的输出是

a is a
b is b

虽然我们有

a Found in --> 
a  is a 
b Found in --> 
b is b
b Found in --> 
b is not f
a Found in --> 
a  is a 
a Found in --> 
a is not e
b Found in --> 
b is b
a Found in --> 
a  is a 
a Found in --> 
a is not e
b Found in --> 
b is b
b Found in --> 
b is not f

我已经尝试了几种方法来实现循环，但没有用

Answer 1

如果我理解正确，您只希望“行”的值为keywords，而不是keywords2。在这种情况下，只需：

for line in a:
    if [x for x in keywords if x in line] and not [x for x in keywords2 if x in line]:
        print line

[x for x in keywords if x in line]循环遍历keywords中的所有项目，如果它们位于line中，则会将其添加到列表中。

[x for x in keywords2 if x in line]循环遍历keywords2中的所有项目，如果它们位于line中，则会将其添加到列表中。

空列表被视为False，您希望keywords列表不为空，keywords2列表为空。

Answer 2

@joon以你的问题风格回答答案;这里还有其他一些想法：

首先遍历行，因为代码应该像“我应该打印此行吗？”这样的问题。对于每一行。

keywords = ("a","b","c","d")
keywords2 = ("e","f","g")

a = ("a  is a ", "a is not e", "b is b", "b is not f")

for line in a:
    for k in keywords:
        if k in line:
            for k2 in keywords2:
                if k2 in line:
                    break
            else:
                print(line)

在线试用：https://repl.it/CgW8

For / else是一个Python的想法，但它有点奇怪 - 如果循环正常结束，else:将运行并打印该行。如果循环中断，else:将不会运行，也不会打印该行。

但更惯用的答案是使用any()，如果列表中的任何内容为True，则返回True，否则返回False：

keywords =("a","b","c","d")
keywords2 = ("e","f","g")
a = ("a  is a ", "a is not e","b is b", "b is not f")

for line in a:
    if any(k in line for k in keywords):
        if not any(k2 in line for k2 in keywords2):
            print(line)

在线试用：https://repl.it/CgW9

Answer 3

这对你有用吗？

keywords =("a","b","c","d")
keywords2 = ("e","f","g")
lines = ("a  is a ", "a is not e","b is b", "b is not f")

for keyword in keywords:
    for line in lines:
        if keyword in line:
            keyword2_exist = False
            for keyword2 in keywords2:
                if keyword2 in line:
                    keyword2_exist = True

            if not keyword2_exist:
                print(line)

Answer 4

这可能对您更有效：

import re

keywords =("a","b","c","d")
keywords2 = ("e","f","g")

pattern = re.compile(r'^(\w+)\s+is\s+(?:not\s+)?(\w+)')
for line in ("a  is a", "a is not e","b is b", "b is not f"):
    for key1, key2 in pattern.findall(line):
        if key1 in keywords and key2 not in keywords2:
            print (line)

根据我的理解，字符串中有两个关键字;每个格式为key is [not] key2，您希望仅打印key位于keywords且key2不在keywords2的行。这基本上就是上述解决方案的作用

Answer 5

你应该学习一些函数式编程，试试这个：

keywords =("a","b","c","d")
keywords2 = ("e","f","g")
a = ("a  is a ", "a is not e","b is b", "b is not f")

for line in a:
    if not any(map(lambda x: x in line, keywords2)):
        if any(map(lambda x: x in line, keywords)):
            print line

Python;如果行中的单词和单词2不一致

5 个答案: