Question

我在列表中有成千上万个句子元素。但是，许多句子不可读。因此，我想在替换过程中忽略这些句子。幸运的是，这些句子具有相似的模式。

已更新

这是我的示例列表：

sentence = ['"//# sourceMappingURL=data:application/json;charset=utf-8;base64,eyJ2ZXJzaW9uIjozLCJzb\n',
        '"// How are you? \n',
        '//# Hello world \n',
        '// sourceMappingURL=data:application/json;charset=utf-8;base64,asdhhi3242jlfaoiuq3r'
       ]

我想在所有句子中用单个空格/代替#，\n，' '等符号，但不可读的元素除外。不可读元素始终具有包含sourceMappingURL=data:application/json;charset=utf-8;base64的模式。

这是我使用Python 3的代码

text = []
for i in sentence:
    isbase64 = "sourceMappingURL=data:application/json;charset=utf-8;base64"
    if i.find(isbase64):
        pass
    else:
        word = i.replace("\n"," ").replace("/"," ").replace("#"," ")
        s = ' '.join(word)
        text.append(s)

但是输出是empty list。我有什么需要改变的吗？

预期输出：

text = ['How are you?','Hello world']

谢谢您的帮助。

Answer 1

使用列表理解：

sentence = ['"//# sourceMappingURL=data:application/json;charset=utf-8;base64,eyJ2ZXJzaW9uIjozLCJzb\n',
        '"// How are you? \n',
        '//# Hello world \n',
        '// sourceMappingURL=data:application/json;charset=utf-8;base64,asdhhi3242jlfaoiuq3r'
       ]

print([x.replace("\n","").replace("/","").replace("#","").strip() for x in sentence if 'sourceMappingURL=data:application/json;charset=utf-8;base64' not in x])

以上代码是以下内容的全面版本：

text = [] 
for x in sentence:
    if 'sourceMappingURL=data:application/json;charset=utf-8;base64' not in x:
        text.append(x.replace("\n","").replace("/","").replace("#","").strip())

print(text)

Answer 2

您的代码看起来大部分都可以正常工作-您刚刚输入了错字！当句子部分包含isbase64时，您用文本...charset:utf...定义了...charset=utf...。

Answer 3

您输入错了：

sourceMappingURL=data:application/json;charset:utf-8;base64

应该是

sourceMappingURL=data:application/json;charset=utf-8;base64和 =

sentence = ['"//# sourceMappingURL=data:application/json;charset=utf-8;base64,eyJ2ZXJzaW9uIjozLCJzb\n',
        '"// How are you? \n',
        '//# Hello world \n',
        '// sourceMappingURL=data:application/json;charset=utf-8;base64,asdhhi3242jlfaoiuq3r'
       ]
text = []
isbase64 = "sourceMappingURL=data:application/json;charset=utf-8;base64"
for i in sentence:
    if isbase64 in i:
        pass
    else:
        word = i.replace("\n"," ").replace("/"," ").replace("#"," ")
        s = ' '.join(word)
        text.append(s)

print text

并使用更多的pythonic方法if string in otherString

干杯：）

Answer 4

i.find(isbase64)返回的是整数而不是布尔值。

如果我确实包含isbase64，它将返回位置（例如，在第一种情况下为5），否则返回-1（例如，在第二个字符串中）。

如果将if替换为：i.find(isbase64)!=-1，它将正常工作（或在i中使用isbase64）。

编辑：您还需要用isbase64 = "sourceMappingURL=data:application/json;charset=utf-8;base64"将isbase64 =替换为该行（您在那里有错字）

Edit2：您还需要打印text才能查看结果，即print(text)

仅在列表中替换特定元素的符号

4 个答案: