假设文本文件file
包含多个离散数字范围,每行一个。每个范围前面都有一个字符串(即范围名称)。每个范围的下限和上限由短划线分隔。每个数字范围都以分号结束。对各个范围进行排序(即,范围101-297在1299-1301之前)并且不重叠。
$cat file
foo 101-297;
bar 1299-1301;
baz 1314-5266;
请注意,在上面的示例中,三个范围不会形成从整数1开始的连续范围。
我认为 awk 是填充缺失数字范围的合适工具,因此所有范围一起形成从{1}到{最后一个范围的上限}的连续范围。如果是这样,你将用什么awk命令/函数来执行任务?
$cat file | sought_awk_command
new1 1-100;
foo 101-297;
new2 298-1298;
bar 1299-1301;
new3 1302-1313;
baz 1314-5266;
-
编辑1 :仔细评估后,下面建议的代码在另一个简单示例中失败。
$cat example2
foo 101-297;
bar 1299-1301;
baz 1302-1314; # Notice that ranges "bar" and "baz" are continuous to one another
qux 1399-5266;
$ awk -F'[ -]' '$3-Q>1{print "new"++o,Q+1"-"$3-1";";Q=$4} 1' example2
new1 1-100;
foo 101-297;
new2 298-1298;
bar 1299-1301;
baz 1302-1314;
new3 1302-1398; # ERROR HERE: Notice that range "new3" has a lower bound that is equal to upper bound of "bar", not of "baz".
qux 1399-5266;
-
编辑2:非常感谢RavinderSingh13帮助解决此问题。但是,建议的代码仍会生成与给定目标不一致的输出。
$ cat example3
foo 35025-35144;
bar 35259-35375;
baz 35376-35624;
qux 37911-39434;
$ awk -F'[ -]' '$3-Q+0>=1{print "new"++o,Q+1"-"$3-1";";Q=$4} {Q=$4;print}' example3
new1 1-35024;
foo 35025-35144;
new2 35145-35258;
bar 35259-35375;
new3 35376-35375; # ERROR HERE: Notice that range "new3" has been added, even though ranges "bar" and "baz" are contiguous.
baz 35376-35624;
new4 35625-37910;
qux 37911-39434;
答案 0 :(得分:2)
尝试:
awk -F'[ -]' '$3-Q>1{print "new"++o,Q+1"-"$3-1";";Q=$4} 1' Input_file
编辑:现在添加一个非单一的线性解决方案,并给出正确的解释。
awk -F'[ -]' ' ###Setting field separator as space, dash here.
$3-Q>1{ ###Checking here if 3rd field and variable Qs subtraction is greater than 1, if yes then perform following.
print "new"++o,Q+1"-"$3-1";"; ###printing the string new with a incrementing value of variable o each time, then variable Qs value with adding 1 to it, then current line $4-1 and semi colon.
Q=$4 ###Assigning the variable Q value to 4th field of the current line here too.
}
1 ###printing the current line here.
' Input_file ###Mentioning the Input_file here too.
EDIT2:根据OP的条件再添加一个答案。
awk -F'[ -]' '$3-Q+0>=1{print "new"++o,Q+1"-"$3-1";";Q=$4} {Q=$4;print}' Input_file
答案 1 :(得分:0)
如原始示例2中bar 1299-1301;
和baz 1301-1314;
在1301
重叠时所显示的范围可以重叠,这没有问题。
$ cat tst.awk
{ split($2,curr,/[-;]/); currStart=curr[1]; currEnd=curr[2] }
currStart > (prevEnd+1) { print "new"++cnt, prevEnd+1 "-" currStart-1 ";" }
{ print; prevEnd=currEnd }
$ awk -f tst.awk file
new1 1-100;
foo 101-297;
new2 298-1298;
bar 1299-1301;
new3 1302-1313;
baz 1314-5266;
$ awk -f tst.awk example2
new1 1-100;
foo 101-297;
new2 298-1298;
bar 1299-1301;
baz 1301-1314;
new3 1315-1398;
qux 1399-5266;
$ awk -f tst.awk example3
new1 1-35024;
foo 35025-35144;
new2 35145-35258;
bar 35259-35375;
baz 35376-35624;
new3 35625-37910;
qux 37911-39434;
答案 2 :(得分:0)
$ cat file1
foo 2-100
bar 102-200
$ awk F' +|[-;}' 'p+1<$2{print "new" ++q, p+1 "-" $2-1 ";"}p=$3' file1
new1 1-1;
foo 2-100
new2 101-101;
bar 102-200
$ cat file2
foo 101-297;
bar 1299-1301;
baz 1314-5266;
$ awk -F' +|[-;]' 'p+1<$2{print "new" ++q, p+1 "-" $2-1 ";"}p=$3' file2
new1 1-100;
foo 101-297;
new2 298-1298;
bar 1299-1301;
new3 1302-1313;
baz 1314-5266;
说明:
$ awk -F' +|[-;]' ' # FS is ; - or a bunch of spaces
p+1 < $2 { # if p revious $3+1 is still less than new $2
print "new"++q,p+1 "-" $2-1 ";" # print a "new" line
}
p=$3 # set future p and implicit print of record *
' file2 # * as all values are above 0