String Matlab中如何从混合字符串中提取日期时间

我在Matlab中有一个100K行的字符串向量。每行包括字母、数字和[./@；，]的混合

我需要检测每行中是否存在这些模式：

MM/dd/YYYY HH:mm
MM.dd.YYYY HH:mm
MM/dd/YY HH:mm
MM.dd.YY HH:mm

例如：

"Read back and verified on 1/15/13 1935 CM;"
"Was negative on 02.10.2015 @ 2015;"
"Result came back positive 4.2.2016 0821;"

输出应为（日期时间格式）：

---

您可以使用以下一组正则表达式。我这样对待两位数的年份：

如果年份大于今天的年份减去2000，则假定为

19xx

，否则假定为

20xx

。预计2099年后会出现问题；-）

Edit:我使用了

strings

作为变量，但它也是一个命令。我已将变量名更改为

string\u list

patterns = [ ...
    '(?<month>\d{1,2})/(?<day>\d{1,2})/(?<year>\d{4}) @? ?(?<hours>\d{2})(?<minutes>\d{2})|'   ... % {m}m/{d}d/YYYY {@ }hhmm
    '(?<month>\d{1,2})\.(?<day>\d{1,2})\.(?<year>\d{4}) @? ?(?<hours>\d{2})(?<minutes>\d{2})|' ... % {m}m.{d}d.YYYY {@ }hhmm
    '(?<month>\d{1,2})/(?<day>\d{1,2})/(?<year>\d{2}) @? ?(?<hours>\d{2})(?<minutes>\d{2})|'   ... % {m}m/{d}d/YY {@ }hhmm
    '(?<month>\d{1,2})\.(?<day>\d{1,2})\.(?<year>\d{2}) @? ?(?<hours>\d{2})(?<minutes>\d{2})'  ... % {m}m.{d}d.YY{@ }hhmm
];

string_list = [ ...
    "Read back and verified on 1/15/13 1935 CM;"
    "Was negative on 02.10.2015 @ 2015;"
    "Result came back positive 4.2.2016 0821;"   
    "Some test for a year earlier than 2000 4.2.89 0821;"   
];

matches = regexp(string_list, patterns,'names');

today = datetime('today');
currentyear = today.Year - 2000;

dates = cell(size(matches));
for i = 1:numel(matches)
    year = str2double(matches{i}.year);
    if year < 100 
        if year > currentyear
            year = year + 1900;
        else
            year = year + 2000;
        end
    end
    dates{i} = datetime(year, str2double(matches{i}.month), str2double(matches{i}.day), str2double(matches{i}.hours), str2double(matches{i}.minutes), 0); 
end

---

也许结尾处的

for

-循环可以简化。

首选输出中的第二个

2015

应该是

20:15

（这是一个时间，不是吗？）并且

MM.dd.YYYY HH:MM

中的

YYYY

应该是

yyyyy

，不是吗？

可以签名吗（您在模式中没有提到）出现在任何模式中，或仅出现在带有

MM.dd.YYYY

的模式中？@PatrickHappel@可能出现在任何模式中。如果有“at”而不是@怎么办？我如何为该模式编写模式？重复

模式中的四行，并将
@
替换为
（？：at）？
。确保
模式
中除最后一行以外的所有行都以
|
结尾。
patterns = [ ...
    '(?<month>\d{1,2})/(?<day>\d{1,2})/(?<year>\d{4}) @? ?(?<hours>\d{2})(?<minutes>\d{2})|'   ... % {m}m/{d}d/YYYY {@ }hhmm
    '(?<month>\d{1,2})\.(?<day>\d{1,2})\.(?<year>\d{4}) @? ?(?<hours>\d{2})(?<minutes>\d{2})|' ... % {m}m.{d}d.YYYY {@ }hhmm
    '(?<month>\d{1,2})/(?<day>\d{1,2})/(?<year>\d{2}) @? ?(?<hours>\d{2})(?<minutes>\d{2})|'   ... % {m}m/{d}d/YY {@ }hhmm
    '(?<month>\d{1,2})\.(?<day>\d{1,2})\.(?<year>\d{2}) @? ?(?<hours>\d{2})(?<minutes>\d{2})'  ... % {m}m.{d}d.YY{@ }hhmm
];

string_list = [ ...
    "Read back and verified on 1/15/13 1935 CM;"
    "Was negative on 02.10.2015 @ 2015;"
    "Result came back positive 4.2.2016 0821;"   
    "Some test for a year earlier than 2000 4.2.89 0821;"   
];

matches = regexp(string_list, patterns,'names');

today = datetime('today');
currentyear = today.Year - 2000;

dates = cell(size(matches));
for i = 1:numel(matches)
    year = str2double(matches{i}.year);
    if year < 100 
        if year > currentyear
            year = year + 1900;
        else
            year = year + 2000;
        end
    end
    dates{i} = datetime(year, str2double(matches{i}.month), str2double(matches{i}.day), str2double(matches{i}.hours), str2double(matches{i}.minutes), 0); 
end

dates =

  4×1 cell array

    {[15-Jan-2013 19:35:00]}
    {[10-Feb-2015 20:15:00]}
    {[02-Apr-2016 08:21:00]}
    {[02-Apr-1989 08:21:00]}