I'm trying to find out if there's a more compact way to write the following format tcsh code, ideally still in tcsh. I spent a while searching around using different keywords but couldn't find anything helpful.
Essentially the code snippet is just looping though a number sequence, and then if that looped value is within a set of numbers, assigning a variable. For simplicity I've made the lists of numbers $VAR is being compared against relatively short, and the number of comparison occurrences few, but the actual problem is double the size is both respects.
foreach VAR (`seq 1 24`)
if ($VAR == 1 || $VAR == 2 || $VAR == 3 || $VAR == 4) then
set cat = small
else if ($VAR == 5 || $VAR == 6 || $VAR == 7 || $VAR == 8) then
set cat = medium
else
set cat = large
endif
end
I suppose I was thinking more along the lines of python where you can just say "if x in [...]" etc., rather than needing to compare $VAR to every number individually as is the case above. I'd considered the following type of setup, but one ends up with more lines overall.
foreach VAR (`seq 1 24`)
foreach C (1 2 3 4)
if ($C == $VAR) then
set cat = small
endif
end
...
end
If the provided code is as simple as is gets in tcsh, is there a more succinct way in say, bash? Thanks for any tips.
I am afraid that in tcsh You won't save any lines thou:
#/bin/tcsh
foreach VAR (`seq 1 24`)
if (${VAR} < 5) then
set cat = "small"
else if (${VAR} > 8) then
set cat = "large"
else
set cat = "medium"
endif
end
exit(0)
You can also consider passing logic to AWK (I do not know if helps in Your case):
#/bin/tcsh
foreach VAR (`seq 1 24`)
set cat = (`echo ${VAR} | awk '$0<5 {print("small"); exit;} $0>8 {print("large"); exit;} {print("medium");}'`)
end
exit (0)
Related
#!/bin/bash
a=0
b=1
echo "give a number:"
read n
clear
echo "the fibonacci sequence until $n:"
for (( i=0; i<n; i++ ))
do
echo -n "$a "
c=$((a + b))
a=$b
b=$c
done
If I interpret it well, this code echoes a $a value after every i++ jumps, then switches the variables as you can see, then on the next i++ loop jump it happens again until "i" reaches "n".
Question: if we want in every loop jump the value of the new "c" why shall we echo $a? I see the connection that: a=$b, b=$c, and c=$((a + b)) but i don't get it why do we refer to $a when doing echo?
Is there a more elegant solution?
You mean, “never ever calculate anything needlessly, ever”? It is possible, of course, but it depends on how much ugliness in the control logic you are willing to tolerate. In the example below, fibonacci1 calculates at most one extra element of the series that may not get printed out and fibonacci2 never calculates any extra series elements and everything makes it to the standard output.
Is any of that “elegant”? Probably not. This is actually a common problem most people encounter when coding (in languages other than purely functional ones): Most high(er)-level languages (unlike e.g. assemblers) provide predefined control structures that work great in typical and obvious cases (e.g. one control variable and one operation per iteration) but may become “suboptimal” in more complex scenarios.
A notoriously common example is a variable that stores a value from the previous iteration. Let’s assume you assign it at the very end of the loop. That works fine, but… Could you avoid the very last assignment (because it is useless), instead of leaving it to the compiler’s wisdom? Yes, you could, but then (e.g.) for ((init; condition; step)); do ...; ((previous = current)); done becomes (e.g.) for ((init;;)); do ...; ((step)); ((condition)) || break; ((previous = current)); done.
On one hand, a tiny bit of something (such as thin air) may have been “saved”. On the other hand, the code became assembler-like and harder to write, read and maintain.
To find a balance there^^^ and {not,} optimize when it {doesn’t,does} matter is a lifelong struggle. It may be something like CDO, which is like OCD, but sorted correctly.
fibonacci1() {
local -ai fib=(0 1)
local -i i
for ((i = $1; i > 2; i -= 2)) {
printf '%d %d ' "${fib[#]}"
fib=($((fib[0] + fib[1])) $((fib[0] + 2 * fib[1])))
}
echo "${fib[#]::i}"
}
fibonacci2() {
trap 'trap - return; echo' return
local -i a=0 b=1 i="$1"
((i)) || return 0
printf '%d' "$a"
((--i)) || return 0
printf ' %d' "$b"
for ((;;)); do
((--i)) || return 0
printf ' %d' "$((a += b))"
((--i)) || return 0
printf ' %d' "$((b += a))"
done
}
for ((i = 0; i <= 30; ++i)); do
for fibonacci in fibonacci{1,2}; do
echo -n "${fibonacci}(${i}): "
"$fibonacci" "$i"
done
done
Working on a little script which put random numbers in a 10 000 size array and then sort all this array with the method ask during the course.
I've done this code but it seem that it begin to sort (when I test I have some "a" that are printed but not as much as supposed to and I don't understand why)
I'm believing the problem come fromes my test on val array, and it's probably a beginner error but I don't really know how to find the problem on th web as I don't really now which line is the problem.
I don't necessary need an answer, just some clues to find it could be good :)
Here is my code: (new to stackoverflow so I don't know how to put a good code view directly, if anyone can show me)
for i in `seq 1 10000`;
do
val[${i}]=$RANDOM
done
echo `date +"%M.%S.%3N"`
FLAG=0
until [ $FLAG -eq 1 ]
do
FLAG=1
for j in `seq 1 9999`;
do
if [ ${val[${j}]} -gt ${val[${j+1}]} ]
then
TMP=${val[${j}]}
val[${j}]=${val[${j+1}]}
val[${j+1}]=$TMP
FLAG=0
echo a
fi
done
done
echo `date +"%M.%S.%3N"`
as asked I can't really have a useful output as I just want the date before and after the sort operation. But the sort is just supposed to put values from lower to higher by taking them two by two and invert them if necessary. Doing this until no numbers are inverted.
Edit: I tried with manual number:
10 3 6 9 1
when running it by putting echo ${val[*]} in the for loop it just print 4 times the same list in the same order, so I'm guessing it doesn't work at all... Is my use of "if" wrong ?
Edit 2: At the begining, I did it in C# and I wanted to do it in shell then, firstly because I wanted to practice shell and then because I wanted to compare efficiency and time needed for the same thing. here is the C# code, working.
Random random = new Random();
int[] _tab = new int[100000];
for (int i = 0; i < _tab.Length; i++)
{
_tab[i] = random.Next(1, _tab.Length);
}
bool perm;
int tmp;
DateTime dt = DateTime.Now;
do
{
perm = false;
for (int i = 0; i < (_tab.Length - 1); i++)
{
if (_tab[i] > _tab[i + 1])
{
tmp = _tab[i];
_tab[i] = _tab[i + 1];
_tab[i + 1] = tmp;
perm = true;
}
}
}
while (perm == true);
Console.WriteLine((DateTime.Now - dt).TotalMilliseconds);
Console.Read();
Thanks :)
If my understanding that you want to know why this script is not producing an "a" indicating the ordering of the array of the numbers initially produced in the "for" loop is correct, then here is a solution:
The syntax is incorrect for your variable expansion. The ${var} cannot have math operators inside the braces, because they have different meaning here. In a normal non-associative array Zsh handles subscripts with some basic math support, so you can use ${array[var+1]} instead of ${array[${var+1}]} as you previously did.
I suspect the reason this came about - complicated, error prone POSIX syntax - would have been avoided by using simplified Zsh syntax, but as stated in an earlier comment, it would not be portable to other shells.
Some shells support similar features: Bash supports most, but not bare subscripts ($array[var]). Strings may be ordered in Zsh in a similar manner, but the math-context brackets (( and )) would have to be replaced with normal test brackets [[ and ]] and the array $val might have to be defined with special typeset options to make the strings compare in the desired manner; that is, they might have to be padded and right or left aligned. For comparing enumeration types, like Jan - Feb, it gets a little more complicated with associative arrays and case-conversion.
Here is the script with the appropriate changes, then again in simplified Zsh:
#!/bin/sh
for i in `seq 1 10000`;
do
val[$((i))]=$RANDOM
done
echo `date +"%M.%S.%3N"`
FLAG=0
until [ $FLAG -eq 1 ]
do
FLAG=1
for j in `seq 1 9999`;
do
if [ ${val[$((j))]} -gt ${val[$((j+1))]} ]
then
TMP=${val[$((j))]}
val[$((j))]=${val[$((j+1))]}
val[$((j+1))]=$TMP
FLAG=0
echo a
fi
done
done
echo `date +"%M.%S.%3N"`
Zsh:
#!/bin/zsh
foreach i ( {1..10000} )
val[i]=$RANDOM
end
echo `date +"%M.%S.%3N"`
FLAG=0
until ((FLAG))
do
FLAG=1
foreach j ( {1..9999} )
if (( val[j] > val[j+1] ))
then
TMP=$val[j]
val[j]=$val[j+1]
val[j+1]=$TMP
FLAG=0
echo a
fi
end
done
echo `date +"%M.%S.%3N"`
Some of you are probably familiar with Project Euler, and I'm currently attempting a few of their problems to teach myself some more bash. They're a bit more mathematical than 'script-y' but it helps with syntax etc.
The problem currently asks me to solve:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
The code I have looks like so:
#!/bin/bash
i="1"
for i in `seq 1 333`
do
threes[$i]=`calc $i*3` # where 'calc' is a function written in bashrc
#calc actually looks like: calc() {awk "BEGIN { print "$*"} }
let "sumthrees = sumthrees + ${threes[$i]}"
done
for i in `seq 1 199`
do
fives[$i]=`calc $i*5`
let "sumfives = sumfives + ${fives[$i]}"
done
let "ans = $sumfives + $sumthrees"
echo "The sum of all 3 factors is $sumthrees and the sum of all five factors is $sumfives"
echo "The sum of both is $ans"
#So I can repeatedly run the script without bash remembering the variables between executions
unset i
unset fives
unset threes
unset sumfives
unset sumthrees
unset ans
So far I've not gotten the correct answer, but have run out of ideas as to where I'm going wrong. (FYI, the script currently gives me 266333, which I believe is close, but I don't know the answer yet.)
Can anyone spot anything? And for my own learning, if there are more elegant solutions to this that people might like to share that would be great.
EDIT
Thanks for all the answers, super informative. Since there are so many useful answers here I'll accept my favourite as the proper thread answer.
Blue Moon pointed out the actual problem with your logic.
You don't need to store all the threes and fives in arrays because you don't need them later.
You don't need to unset variables at the end of a script if you use ./yourscript or bash script because they'll disappear along with the
shell instance (better to initialize them first in any case).
You don't need awk to do math, bash does that just fine.
seq and let are not the best way to do anything in a bash script.
Here's a straight forward version:
#!/bin/bash
sum=0
for ((i=1; i<1000; i++))
do
if (( i%3 == 0 || i%5 == 0 ))
then
(( sum += i ))
fi
done
echo "$sum"
Your logic is almost right except that there are numbers which divide by both 3 and 5. So you are adding these numbers twice. Hence, you get wrong answer.
Use another loop similar to ones you have and subtract the ones that divide by both 3 and 5 from the result.
A few tips you might find useful:
In bash, you use let to give the shell a hint that a variable should be considered a number. All bash variables are strings, but you can do arithmetic on numerical strings. If I say let i=1 then i is set to 1, but if I say let i="taco" then $i will be 0, because it couldn't be read as a number. You can achieve a small amount of type-safety when doing mathematical work in the shell.
Bash also has $((this)) mechanism for doing math! You can check it out yourself: echo $((2 + 2)) -> 4, and even more relevant to this problem: echo $((6 % 3 == 0)) -> 1
In case you aren't familiar, % divides the first number by the second, and gives back the remainder; when the remainder is 0, it means that the first is divisible by the second! == is a test to see if two things are equal, and for logical tests like this 1 represents true and 0 represents false. So I'm testing if 6 is divisible by 3, which it is, and the value I get back is 1.
The test brackets, [ ... ] have a "test for equality" flag, -eq, which you can use to check if a math expression has a certain value (man test for more details)!
$ let i=6
$ echo $((i % 3 == 0 || i % 5 == 0))
1
$ if [ $((i % 3 == 0 || i % 5 == 0)) -eq 1 ]; then echo "yes"; fi
yes
(|| is another logical test - $((a || b)) will be 1 (true) when a is true or b is true).
Finally, instead of doing this for the number 6, you could do it in a for loop and increment a sum variable every time you find a multiple of 3 or 5:
let sum=0
for i in {1..1000}; do
if [ $((i % 3 == 0 || i % 5 == 0)) -eq 1 ]; then
let sum=$((sum + i))
fi
done
echo $sum
And there you'd have a working solution!
Bash has a lot of nice little tricks, (and a lot more mean ugly tricks), but it's worth learning at least a handful of them to make use of it as a scripting tool.
How about creative use of the modulus function & some checks. Then you have just 1 loop.
#!/bin/bash
i=1
while [ $i -lt 1000 ]
do
if [ $(($i % 3)) -eq 0 ] || [ $(($i % 5)) -eq 0 ]
then
sumall=$(($sumall+$i))
fi
i=$(($i+1))
done
echo "The sum of both is $sumall"
Answer: 233168
A different solution:
#!/bin/bash
sum=0
for n in {1..999}; do [ $(((n%5) * (n%3))) -eq 0 ] && sum=$((sum+n)); done
echo $sum
The script loops through all numbers below 1000, tests if the product of the number mod 3 and the number mod 5 is 0 (the product of two numbers can only be zero if one of them is zero). If that is the case, it adds the current number to a sum, which is printed out afterwards.
By the way, if I were you I'd include the definition of the calc function inside the script, to get a self-contained solution that doesn't need your specific configuration.
I have a file which contains a list of numbers defined as follow :
var1=0x00000001
var2=0x00000002
var3=0x00000008
var4=0x00000020
var5=0x00000040
var6=0x00000080
var7=0x00000100
var8=0x00000200
var9=0x00000400
var10=0x00000800
var11=0x000001000
var12=0x000002000
var13=0x000004000
var14=0x000008000
var15=0x00010000
var16=0x00020000
var17=0x00040000
var18=0x10000000
var19=0x20000000
var20=0x40000000
var21=0x80000000
I want to write something like this:
decValue=2147483650
printf -v hexValue "%x" "${decValue}"
echo $hexValue
IFS="="
while read name ID x
do
test $((${hexValue} & ${ID})) = 0 && continue
array+=("${name}")
done < "$FILE_NAME"
It returns :
80000002
var2 var9 var11 var12 var14 var17
But, in this specific case I just would like to return :
var21 var2
Other example, if decValue=12288 I would like to return var11 and var12.
Bitwise operators is a good tool to solve this issue ?
Use
printf -v hexValue "%#x" "${decValue}"
(or use ${decimalValue} in the test inside the loop)
As it is now, $hexValue ends up being 80000002 (as your own echo statement shows), and this is later interpreted as a decimal number when you want it to be interpreted as a hexadeximal one.
Passing %#x as format specifier to printf will make $hexValue have the value 0x80000002.
You'll also have to take another good look at the var table; there are a number of gaps in it. 0x4 is missing between var2 and var3, 0x10 is missing between var3 and var4, and between var17 and var18, the whole block from 0x80000 to 0x8000000 is gone as well. You're not going to get the results you expect for variables that have any of these bits set.
It might also be worth a thought to generate the bitmasks on the fly rather than holding them precalcuated in a file. One possible approach for that is
for((i = 0; (1 << i) <= $hexValue; ++i))
do
test $(($hexValue & (1 << i))) = 0 && continue
# Note: this will remember (zero-based) bit numbers rather than variable
# names because there are no named variables any longer
array+=($i)
done
In this, the bitshift expression 1 << i gives the number 2i, or put another way: 1 << i has the same value as var$((i + 1)) would in a repaired table.
Or use perl as:
perl -F= -slnE 'say $_ if( hex($F[1]) & $num )' -- -num=12288 < file
prints:
var11=0x000001000
var12=0x000002000
or
perl -F= -slnE 'say $F[0] if( hex($F[1]) & $num )' -- -num=12288 < file
prints
var11
var12
input from file $2 : 1 -> 2
while read -a line; do
if (( ${line[2]} > linesNumber )); then
echo "Graph does not match known sites4"
exit
fi
done < "$2"
For some reason inside the if condition, the value of ${line[2]) is not 2
but if I print the value outside if:
echo `${line[2]}`
2
What's linesNumber? Even if you put $linesNumber, where is it coming from?
If you are tracking the line number, you need to set it and increment it. Here's my sample program and data. It's inspired by your example, but doesn't do exactly what you want. However, it shows you how to setup a variable that tracks the line number, how to increment it, and how to use it in an if statement:
foo.txt:
this 1
that 2
foo 4
barf 4
flux 5
The Program:
lineNum=0
while read -a line
do
((lineNum++))
if (( ${line[1]} > $lineNum ))
then
echo "Line Number Too High!"
fi
echo "Verb = ${line[0]} Number = ${line[1]}"
done < foo.txt
Output:
Verb = this Number = 1
Verb = that Number = 2
Line Number Too High!
Verb = foo Number = 4
Verb = barf Number = 4
Verb = flux Number = 5