我有这个问题,如果它包含'非法'字符,我想删除列表元素。合法字符在多个列表中指定。它们是这样形成的,<body>
代表字母表(az + AZ),$.fn.addLoader = function (loadText: string): JQuery {
// prepare divs as constants
const overlayDiv = $("<div class=\"load-overlay\" />");
const spinnerDiv = $("<div class=\"load-spinner\" />"); // Note here, you can self-close elements created by JQuery if they're empty.
//overlayDiv.append(spinnerDiv); // I've removed this from here as you don't need to append the spinner to the overlay.
if (loadText) {
const textDiv = $("<div class=\"load-text\">" + loadText + "</div>");
overlayDiv.append(textDiv);
}
// once this is all done, you then append the spinner to the body (or the parent container, whichever is preferable)
$('body').append(spinnerDiv);
overlayDiv.appendTo(this.css("position", "relative"));
return this;
代表数字(0-9),alpha
代表标点符号(排序)。< / p>
digit
这样我可以将某些内容指定为非法字符,如果它没有出现在其中一个列表中。
之后我有一个包含元素的列表:
punct
我想过滤掉包含非法字符的元素。所以这是我想得到的结果(不需要订购):
alpha = list(string.ascii_letters)
digit = list(string.digits)
punct = list(string.punctuation)
编辑:
我试过(以及它的所有变体):
Input = ["Amuu2", "Q1BFt", "dUM€n", "o°8o1G", "mgF)`", "ZR°p", "Y9^^M", "W0PD7"]
其中仅使用过滤后的元素创建新列表,但问题是当包含至少一个合法字符时添加元素。例如:var = ["Amuu2", "Q1BFt", "mgF)`", "Y9^^M", "W0PD7"]
已添加,因为它包含Z,R和p。
我也尝试过:
for InItem in Input:
if any(AlItem in InItem for AlItem in alpha+digit+punct):
FilInput.append(InItem)
但在那之后,我无法弄清楚如何删除元素。 哦,还有一点小小的提示,为了让它变得更加困难,如果它有点快,那将会很好,因为它需要做数百万次。但它需要首先工作。
答案 0 :(得分:5)
定义一组合法字符。然后应用列表理解。
>>> allowed = set(string.ascii_letters + string.digits + string.punctuation)
>>> inp = ["Amuu2", "Q1BFt", "dUM€n", "o°8o1G", "mgF)`", "ZR°p", "Y9^^M", "W0PD7"]
>>> [x for x in inp if all(c in allowed for c in x)]
['Amuu2', 'Q1BFt', 'mgF)`', 'Y9^^M', 'W0PD7']
答案 1 :(得分:1)
您可以使用列表理解,如果所有字符都符合您的条件,请与all
核对:
>>> [element for element in Input if all(c in alpha + digit + punct for c in element)]
['Amuu2', 'Q1BFt', 'mgF)`', 'Y9^^M', 'W0PD7']
答案 2 :(得分:1)
正如您所提到的,只要any
字符是正确的,就会附加字词。您需要检查它们是否all
正确无误:
filtered_words = []
for word in words:
if all(char in alpha+digit+punct for char in word):
filtered_words.append(word)
print(filtered_words)
# ['Amuu2', 'Q1BFt', 'mgF)`', 'Y9^^M', 'W0PD7']
您还可以检查是否有一个不正确的字符:
filtered_words = []
for word in words:
if not any(char not in alpha+digit+punct for char in word):
filtered_words.append(word)
print(filtered_words)
它的可读性要低得多。
为了提高效率,您不应在每次迭代期间使用alpha+digit+punct
连接列表。你应该在任何循环之前一劳永逸地做到这一点。从这些列表中创建一个集合也是一个好主意,因为当有许多允许的字符时,char in set
比char in list
快得多。
最后,您可以使用列表推导来避免for循环。如果您这样做,最终会得到@timgeb's solution:)
您可以从列表中创建正则表达式模式,并查看匹配的字词:
# encoding: utf-8
import string
import re
alpha = list(string.ascii_letters)
digit = list(string.digits)
punct = list(string.punctuation)
words = ["Amuu2", "Q1BFt", "dUM€n", "o°8o1G", "mgF)`", "ZR°p", "Y9^^M", "W0PD7"]
allowed_pattern = re.compile(
'^[' +
''.join(
re.escape(char) for char in (
alpha +
digit +
punct)) +
']+$')
# ^[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789\!\"\#\$\%\&\'\(\)\*\+\,\-\.\/\:\;\<\=\>\?\@\[\\\]\^_\`\{\|\}\~]+$
print([word for word in words if allowed_pattern.match(word)])
# ['Amuu2', 'Q1BFt', 'mgF)`', 'Y9^^M', 'W0PD7']
你也可以写:
print(list(filter(allowed_pattern.match, words)))
# ['Amuu2', 'Q1BFt', 'mgF)`', 'Y9^^M', 'W0PD7']
re.compile
可能比简单地初始化set
需要更多时间,但过滤速度可能会更快。
答案 3 :(得分:1)
对于您的问题,这是一个“非”有效的解决方案,但学习如何循环列表,字符等可能很有趣。
# coding=utf-8
import string
# Aux var
result =[]
new_elem = ""
# lists with legal characters
alpha = list(string.ascii_letters)
digit = list(string.digits)
punct = list(string.punctuation)
# Input strings
Input = ["Amuu2", "Q1BFt", "dUM€n", "o°8o1G", "mgF)`", "ZR°p", "Y9^^M", "W0PD7"]
# Loop all elements of the list and each char of them
for elem in Input:
## check each char
for char in elem:
if char in alpha:
#print 'is ascii'
new_elem += char
elif char in digit:
#print 'is digit'
new_elem += char
elif char in punct:
#print 'is punct'
new_elem += char
else:
new_elem = ""
break
## Add to result list
if new_elem != "":
result.append(new_elem)
new_elem = ""
print result