假设我有一个包含 -用逗号分隔的字符串 -和文字
my_string = "2 Marine Cargo 14,642 10,528 16,016 more text 8,609 argA 2,106 argB"
我想将它们提取到一个数组中,该数组由“数字”和“单词组”分开
resultArray = {"2", "Marine Cargo", "14,642", "10,528", "16,016",
"more text", "8,609", "argA", "2,106", "argB"};
注释0:每个条目之间可能有多个空格,应将其忽略。
注释1:“船用货物”和“更多文本”没有分成不同的字符串,因为它们是一组单词,没有数字将它们分开。 而argA和argB是分开的,因为它们之间存在数字。
答案 0 :(得分:3)
您可以尝试使用此正则表达式进行拆分
([\d,]+|[a-zA-Z]+ *[a-zA-Z]*) //note the spacing between + and *.
[a-zA-Z] + [a-zA-Z] //将搜索一个单词,然后搜索空格(如果有),然后搜索另一个单词(如果任何)。
String regEx = "[0-9,]+|[a-zA-Z]+ *[a-zA-Z]*";
您这样使用它们
public static void main(String args[]) {
String input = new String("2 Marine Cargo 14,642 10,528 16,016 more text 8,609 argA 2,106 argB");
System.out.println("Return Value :" );
Pattern pattern = Pattern.compile("[0-9,]+|[a-zA-Z]+ *[a-zA-Z]*");
ArrayList<String> result = new ArrayList<String>();
Matcher m = pattern.matcher(input);
while (m.find()) {
System.out.println(">"+m.group(0)+"<");
result.add(m.group(0));
}
}
以下是从https://regex101.com自动生成的RegEx的输出以及详细说明。
1st Alternative [0-9,]+
Match a single character present in the list below [0-9,]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
0-9 a single character in the range between 0 (index 48) and 9 (index 57) (case sensitive)
, matches the character , literally (case sensitive)
2nd Alternative [a-zA-Z]+ *[a-zA-Z]*
Match a single character present in the list below [a-zA-Z]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
a-z a single character in the range between a (index 97) and z (index 122) (case sensitive)
A-Z a single character in the range between A (index 65) and Z (index 90) (case sensitive)
* matches the character literally (case sensitive)
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below [a-zA-Z]*
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
a-z a single character in the range between a (index 97) and z (index 122) (case sensitive)
A-Z a single character in the range between A (index 65) and Z (index 90) (case sensitive)
答案 1 :(得分:1)
如果空格是您的问题。 String#split使用正则表达式作为参数。然后,您可以这样做:
my_list = Arrays.asList(my_string.split("\s?"));
但是,这并不能解决所有问题,如评论中提到的那样。
答案 2 :(得分:1)
您可以这样做:
List<String> strings = new ArrayList<>();
String prev = null;
for (String w: my_string.split("\\s+")) {
if (w.matches("\\d+(?:,\\d+)?")) {
if (prev != null) {
strings.add(prev);
prev = null;
}
strings.add(w);
} else if (prev == null) {
prev = w;
} else {
prev += " " + w;
}
}
if (prev != null) {
strings.add(prev);
}
答案 3 :(得分:1)
我喜欢Angel Koh solution,并希望添加它。仅当数字部分由一或两个部分组成时,他的解决方案才会匹配。
如果您还想捕获由三个或更多部分组成的部分,则必须稍微修改一下正则表达式为:([\d,]+|[a-zA-Z]+(?: *[a-zA-Z])*)
非捕获组(?: *[a-zA-Z])
重复无限次(如果需要),并将捕获所有纯数字部分。