我有一个文件,其中包含从某处复制的数字。看起来像这样:
{02 12 04 01 07 10 11 06 08 05 03 15 13 00 14 09},
{14 11 02 12 04 07 13 01 05 00 15 10 03 09 08 06},
{04 02 01 11 10 13 07 08 15 09 12 05 06 03 00 14},
{11 08 12 07 01 14 02 13 06 15 00 09 10 04 05 03}
现在,我必须在每个数字之后添加逗号(基本上是使其成为C ++数组),但是如您所见,这样做可能非常繁琐,特别是如果您有很多的话。
我试着像这样使用sed:
cat file.txt | sed -r "s/ /, /g"
但是,如果我用这个,我将用“,space”替换每个“空格”,而我只想用“,”替换数字后面的空格。
如果使用cat file.txt | sed -r "s/[0123456789] /, /g"
,在更换之前我将无法获得相同的号码。因此,我只想更改子字符串的某些部分。
我该怎么做呢?
cat file.txt | sed -r 's/([0-9]+)/\1,/g'
{02, 12, 04, 01, 07, 10, 11, 06, 08, 05, 03, 15, 13, 00, 14, 09,},
{14, 11, 02, 12, 04, 07, 13, 01, 05, 00, 15, 10, 03, 09, 08, 06,},
{04, 02, 01, 11, 10, 13, 07, 08, 15, 09, 12, 05, 06, 03, 00, 14,},
{11, 08, 12, 07, 01, 14, 02, 13, 06, 15, 00, 09, 10, 04, 05, 03,}
解释:
First capturing group ([0-9]+)
Match a single character (i.e. number) present in the table [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)
In other words, the [0-9]+ pattern matches an integer number (without decimals) even Inside longer strings, even words.
\1 is called a "back reference" or "special escapes" in the sed documentation. It refers to the corresponding matching sub-expressions in the regexp. In other words, in this example, it inserts the contents of each captured number in the table followed by comma.
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