我正在寻找一种从这个字母块中获得对角线组合的方法:
a b c d e f
h i j k l m
o p q r s t
v w x y z a
c d e f g h
j k l m n o
我有这个:awk '{++f; print $(f + 0)}' file.txt
但这只会让我(如果我可以增加,f + 0,不知何故增加1次,6次):
a i q y g o
b j r z h
c k s a
d l t
e m
f
我需要另一半......就像这样(不必按此顺序:
a i q y g o
h p x f n
b j r z h
o w e m
c k s a
v d l
d l t
c k
e m
j
f
答案 0 :(得分:2)
您可以使用GNU datamash:
$ datamash -t' ' transpose < file
a h o v c j
b i p w d k
c j q x e l
d k r y f m
e l s z g n
f m t a h o
答案 1 :(得分:1)
perl -lane'
push @r, [@F];
END {
for my $n (0 .. $#r) {
my (@x,@y);
for (0 .. $#r) {
push @x, $r[$n+$_][$_];
push @y, $r[$_][$n+$_];
}
print "@x";
print "@y" if $n;
}
}
' file
输出
a i q y g o
h p x f n
b j r z h
o w e m
c k s a
v d l
d l t
c k
e m
j
f
答案 2 :(得分:0)
这是阵列的工作。
#!/usr/bin/awk -f
# Print diagonals from a square of text.
# This script has no error checking.
# Written by PM 2Ring 2014.10.13
{
for(i=1; i<=NF; i++)
a[NR,i] = $i
}
END{
for(i=1; i<=NR; i++)
{
for(j=1; i+j-1<=NR; j++)
printf "%s ", a[i+j-1, j]
print ""
if (i == 1)
continue
for(j=1; i+j-1<=NR; j++)
printf "%s ", a[j, i+j-1]
print ""
}
}
这是上述代码的一个版本,用于将每个对角字存储在一个数组中。
#!/usr/bin/awk -f
# Print diagonals from a square of text.
# This script has no error checking.
# Written by PM 2Ring 2014.10.13
{
for(i=1; i<=NF; i++)
a[NR,i] = $i
}
END{
k = 1
for(i=1; i<=NR; i++)
{
word = ""
for(j=1; i+j-1<=NR; j++)
word = word a[i+j-1, j]
words[k++] = word
if (i == 1)
continue
word = ""
for(j=1; i+j-1<=NR; j++)
word = word a[j, i+j-1]
words[k++] = word
}
numwords = k
for (k=1; k<numwords; k++)
print words[k]
}
<强>输出强>
aiqygo
hpxfn
bjrzh
owem
cksa
vdl
dlt
ck
em
j
f
打印正交对角线,例如&#39; flrxdj&#39;,留给读者练习。
...
这是一个打印给定单词中所有子词的函数。
#!/usr/bin/awk -f
# Print subwords of a given word
# Written by PM 2Ring 2014.10.13
function print_subwords(s, i, j, len)
{
len = length(s)
for (i=1; i<=len; i++)
{
print "Length:", i
for (j=1; i+j-1 <= len; j++)
print substr(s, j, i)
}
}
BEGIN{
print_subwords("ABCD")
}
<强>输出强>
Length: 1
A
B
C
D
Length: 2
AB
BC
CD
Length: 3
ABC
BCD
Length: 4
ABCD