Question

以下摘录几乎是我想要的，但不完全是：

echo 'pie(x=x,labels=c("5.3 %","2.8 %","1.5 %","5.2 %","2.9 %","2.3 %","7.9 %","16.9 %","19.5 %","27.5 %","8.3 %"),main="ChIP",col=c("#445FA2","#EB9D86","#799F7A","#6C527F","#5FA1C1","#E8BB77","#A8C5EF","#FDCDB9","#C6E6B5","#F1D5EE","#B4E1F6"),clockwise=TRUE,border=FALSE,radius=0.9,cex=0.8,init.angle=90,density=100)'|grep -oP '(?<=").*?(?= %)'

输出是：

5.3
,"2.8
,"1.5
,"5.2
,"2.9
,"2.3
,"7.9
,"16.9
,"19.5
,"27.5
,"8.3

请您告诉我如何获得一个清洁剂＆＃39;输出如：

5.3
2.8
1.5
5.2
2.9
2.3
7.9
16.9
19.5
27.5
8.3

最好，请在上面的grep -oP上找到解决方案，但当然欢迎其他解决方案！

Answer 1

如果你只想要0-9和＆＃34;。＆＃34;你可以使用的字符：

grep -oP '(?<=")[\d\.]*?(?= %)'

而不是

grep -oP '(?<=").*?(?= %)'

注意：Perl中的\ d匹配0-9而不是。在Perl中匹配任何字符，因此您将其转义为文字句点字符

Answer 2

带空格的简单匹配数字适用于您想要的输出。

 grep -oP '[0-9]+.[0-9]+ '

如果小数点与%之间没有空格，这也可以，

grep -oP '[0-9]+.[0-9]+.*?(?=%)'

Answer 3

用以下内容扩展它：

... | grep -Eo "[0-9]+\.[0-9]+"

如果您只是寻找“数字点数”或使用：

... | grep -Eo "[0-9]+\.[0-9]+|[0-9]+"

所有数字之间可能有一个点。

Answer 4

GNU awk 解决方案：

s='pie(x=x,labels=c("5.3 %","2.8 %","1.5 %","5.2 %","2.9 %","2.3 %","7.9 %","16.9 %","19.5 %","27.5 %","8.3 %"),main="ChIP",col=c("#445FA2","#EB9D86","#799F7A","#6C527F","#5FA1C1","#E8BB77","#A8C5EF","#FDCDB9","#C6E6B5","#F1D5EE","#B4E1F6"),clockwise=TRUE,border=FALSE,radius=0.9,cex=0.8,init.angle=90,density=100)'

awk -v FPAT='"[0-9]+\\.[0-9]+ %"' '{ for(i=1;i<=NF;i++) { gsub(/[^.0-9]+/,"",$i); print $i } }' <<<"$s"

输出：

5.3
2.8
1.5
5.2
2.9
2.3
7.9
16.9
19.5
27.5
8.3

使用grep在两个子字符串之间提取字符串

4 个答案: