SOLVED:
I'm trying to write a script that calculates the distance, bearing, and mid-point given a pair of lat lon coords.
I found the formula's easily enough, but I'm getting the wrong answers. It might just be a math mistake, but I've looked it over multiple times and I'm missing something.
I'm following this website's formulas: http://www.movable-type.co.uk/scripts/latlong-nomodule.html
Here's what I get, output is dist, bearing, midpoint lat, midpoint long
script.bash 1 -80 -3 -79.2
453.58,158.22,68.1258,95.390
this is what I should get:
script.bash 1 -80 -3 -79.2
453.6 168.7 -1 -79.6
So, distance looks good. But the others are all off. Any thoughts?
Here is my code:
#!/bin/bash
lat1=$1
lat2=$3
lon1=$2
lon2=$4
#some basic info
R=6371
lat1r=`echo "$lat1*3.14159/180" | bc -l`
lat2r=`echo "$lat2*3.14159/180" | bc -l`
lon1r=`echo "$lon1*3.14159/180" | bc -l`
lon2r=`echo "$lon2*3.14159/180" | bc -l`
dLat=`echo "$lat2r - $lat1r" | bc -l`
dLon=`echo "$lon2r - $lon1r" | bc -l`
#Distance calculations
a=`echo "-s ($dLat/2) * -s ($dLat/2) + -c ($lat1r) * -c ($lat2r) * -s ($dLon/2) * -s ($dLon/2)" | bc -l`
c1=`echo "sqrt($a) " | bc -l`
c2=`echo "sqrt(1 - $a)" | bc -l`
cat=`echo "$c1,$c2"| awk -F',' '{ print atan2($1,$2) }'`
c=`echo "2*$cat" | bc -l`
d=`echo "$R*$c" | bc -l`
#Bearing calculation
y=`echo "-s ($dLon) * -c ($lat2r)" | bc -l`
x=`echo "-c ($lat1r) * -s ($lat2r) - -s ($lat1r) * -c ($lat2r) * -c ($dLon)" | bc -l`
brng=`echo "$y,$x"| awk -F',' '{ print atan2($1,$2) }'`
brn=`echo "$brng * 180 / 3.14159" | bc -l`
echo "$brng * 180 / 3.14159"
#Mid point calculation
Bx=`echo "-c ($lat2r) * -c ($dLon)" | bc -l`
By=`echo "-c ($lat2r) * -s ($dLon)" | bc -l`
atc1=`echo " -s ($lat1r) + -s ($lat2r)" | bc -l`
atc2=`echo " sqrt( ( -c ($lat1r) + $Bx )^2 + $By^2 ) " | bc -l`
latmidr=`echo "$atc1,$atc2"| awk -F',' '{ print atan2($1,$2) }'`
latmid=`echo "$latmidr * 180 / 3.14159" | bc -l`
atc3=$By
atc4=`echo " -c ($lat1r) + $Bx" | bc -l`
lonmidr=`echo "$atc3,$atc4"| awk -F',' '{ print atan2($1,$2) }'`
lonmid=`echo "$lonmidr * 180 / 3.14159" | bc -l`
echo $d,$brn,$latmidr,$lonmid
That's a completely inappropriate job for a shell script, you should have done it as a single awk (or similar, e.g. perl, ruby, python) script. Btw naming a variable the same as a command (cat) obfuscates your code and makes it more error prone.
Here's what your starting point should be (check the math/conversions as I almost certainly didn't always interpret what you were trying to do when piping strings with -s and -c to bc etc. correctly as I was guessing):
$ cat tst.sh
#!/usr/bin/env bash
awk -v lat1="$1" -v lat2="$3" -v lon1="$2" -v lon2="$4" '
BEGIN {
#some basic info
pi = 3.14159
R = 6371
lat1r = lat1 * pi / 180
lat2r = lat2 * pi / 180
lon1r = lon1 * pi / 180
lon2r = lon2 * pi / 180
dLat = lat2r - lat1r
dLon = lon2r - lon1r
#Distance calculations
a = sin(dLat/2) * sin(dLat/2) + cos(lat1r) * cos(lat2r) * sin(dLon/2) * sin(dLon/2)
c1 = sqrt(a)
c2 = sqrt(1 - a)
cat = atan2(c1,c2)
c = 2 * cat
d = R * c
#Bearing calculation
x = cos(lat1r) * sin(lat2r) - sin(lat1r) * cos(lat2r) * cos(dLon)
y = sin(dLon) * cos(lat2r)
brng = atan2(y,x)
brn = brng * 180 / pi
print brng * 180 / pi
#Mid point calculation
Bx = cos(lat2r) * cos(dLon)
By = cos(lat2r) * sin(dLon)
atc1 = sin(lat1r) + sin(lat2r)
atc2 = sqrt( (cos(lat1r) + Bx )^2 + By^2 )
latmidr = atan2(atc1,atc2)
latmid = latmidr * 180 / pi
atc3 = By
atc4 = cos(lat1r) + Bx
lonmidr = lon1r + atan2(atc3,atc4)
lonmid = lonmidr * 180 / pi
print d, brn, latmid, lonmid
}
'
.
$ ./tst.sh 1 -80 -3 -79.2
168.696
453.581 168.696 -1.00002 -79.6002
Related
I want to create a histogram with total file count intervals of 50 on Y-axis and time created in weeks on X-axis (i.e if new files were created between week 1 and 2 and so on)
Something like
200, 150, 100 , 50 files created during a certain week
7, 14, 21, 28 days on Y-axis. Kind of lost on how to implement this. Any help is appreciated
Update: I am trying along these lines
find <dirname> -type f -ctime -1 -ctime -7 | wc -l
find <dirname> -type f -ctime +7 -ctime -14 | wc -l
Find the max number and use this as my X-axis upper limit. Then divide this number into equal intervals to plot my X-axis
This is a start using GNU awk for time functions (untested since you didn't provide concise, testable sample input that we could test against):
find "$1" -type f -printf '%T# %p\0' |
awk -v RS='\0' '
BEGIN {
nowSecs = systime()
}
{
fileName = gensub(/\S+\s+/,"",1)
fileModSecs = int($1)
fileAgeSecs = nowSecs - fileModSecs
fileAgeDays = int(fileAgeSecs / (24 * 60 * 60))
fileAgeWeeks = int(fileAgeDays / 7)
weekNr = fileAgeWeeks + 1
fileCnts[weekNr]++
numWeeks = (weekNr > numWeeks ? weekNr : numWeeks)
maxFileCnt = (fileCnts[weekNr] > maxFileCnt ? fileCnts[weekNr] : maxFileCnt)
print nowSecs, fileModSecs, fileAgeSecs, fileAgeDays, fileAgeWeeks, weekNr, fileName | "cat>&2"
}
END {
for (fileCnt=maxFileCnt; fileCnt>0; fileCnt--) {
for (weekNr=1; weekNr<=numWeeks; weekNr++) {
if (weekNr in fileCnts) {
char[weekNr] = "*"
}
printf "%s%s", char[weekNr], (weekNr<numWeeks ? OFS : ORS)
}
}
for (weekNr=1; weekNr<=numWeeks; weekNr++) {
printf "%s%s", weekNr, (weekNr<numWeeks ? OFS : ORS)
}
}
'
You need to figure out the details of the loops in the END section for printing the histogram but the above at least shows you how to get the count of files by week without calling find multiple times and hard-coding the number of days week by week.
Apologies being ksh instead of bash (bash level is near echo "Hello World") :)...
Would that do what you need ?
#!/bin/ksh
######################################
#
# statDirReport.sh
#
version="1.0"
# Andre Gelinas, 2018
#
######################################
#############
# Variables
#############
typeset -F2 SCALE
# Max value of X
X_SCALE=30
#############
# Main
#############
if [[ -n $1 ]]; then
DIRNAME=$1
else
print -n "Enter full path to stat : "; read DIRNAME
fi
if [[ ! -d $DIRNAME || ! -r $DIRNAME || ! -x $DIRNAME ]]; then
print "ERROR - Directory unusable - Exiting"
exit
fi
## Getting the data
CTIME1=1
CTIME2=0
for ((i=1;i<=4;i++)); do
CTIME2=$(($i*7))
FILE_COUNT[$i]=$(find $DIRNAME -type f -ctime +$CTIME1 -ctime -$CTIME2 | wc -l)
#To find late on the max amount
F_COUNT[${FILE_COUNT[$i]}]=${FILE_COUNT[$i]}
#
CTIME1=$CTIME2
done
#Doing some math
## Highest number of file
MAX_COUNT=${F_COUNT[-1]}
## Find the value of each tick
SCALE=$(($MAX_COUNT/$X_SCALE))
## Find the real length of the histogram for each week
## having the highest amount using full x scale (integer mathematics)
for ((i=1;i<=4;i++)); do
DATA_2_SCALE[$i]=$(((${FILE_COUNT[$i]}*$X_SCALE)/$MAX_COUNT))
done
# Getting the report
typeset -L2 Col1
typeset -L1 Col2
typeset -L$(($X_SCALE+5)) Col3
typeset -L5 Col4
Col1="Wk"
Col2=" "
Col3="Data"
Col4="Real"
clear
print "statDirReport v$version\tScale is #=$SCALE\n"
print "$Col1$Col2$Col3$Col4\n"
for ((i=1;i<=4;i++)); do
Col1=$i
Col2="|"
graph=""
Col4=${FILE_COUNT[$i]}
for ((j=1;j<=${DATA_2_SCALE[$i]};j++)); do
graph+="#"
done
Col3=$graph
print "$Col1$Col2$Col3$Col4"
done
Edit to modify to add dates as title for the histograms. Modify the last part, right after the "DATA_2_SCALE" loop, with :
#Setting the title of each histogram
## Finding how many sec since the beginning of time
TODAY_SEC=$(date +"%s")
## Finding real date for find range
SEC_PER_DAY=86400
lastDate=$(date -u -d #"$TODAY_SEC" +"%m/%d")
for ((i=1;i<=4;i++)); do
firstDate=$(date -u -d #"$(($TODAY_SEC-(7*$i*$SEC_PER_DAY)))" +"%m/%d")
WEEK[$i]=$firstDate" to "$lastDate" "
lastDate=$firstDate
done
# Getting the report
typeset -L15 Col1
typeset -L1 Col2
typeset -L$(($X_SCALE+5)) Col3
typeset -L5 Col4
Col1="Wk"
Col2=" "
Col3="Data"
Col4="Real"
clear
print "statDirReport v$version\tScale is #=$SCALE\n"
print "$Col1$Col2$Col3$Col4\n"
for ((i=1;i<=4;i++)); do
Col1=${WEEK[$i]}
Col2="|"
graph=""
Col4=${FILE_COUNT[$i]}
for ((j=1;j<=${DATA_2_SCALE[$i]};j++)); do
graph+="#"
done
Col3=$graph
print "$Col1$Col2$Col3$Col4"
done
Using feedgnuplot on a home directory:
dirname=~
e=0
for f in `seq 7 7 28` ; do
find "${dirname}" -type f -ctime +$e -ctime -$f | wc -l
e=$f
done 2> /dev/null |
feedgnuplot --terminal 'dumb 50,15' --with boxes --unset grid --exit
Output:
5500 +-+-----+-------+------+-------+-----+-+
5000 +-+ ********* + + + +-+
4500 +-+ * * ******** +-+
4000 +-+ * * * * +-+
3500 +-+ * * * * +-+
3000 +-+ * * * * +-+
2500 +-+ * ********* * +-+
2000 +-+ * * * * +-+
1500 +-+ * * * * +-+
1000 +-+ * + * + * + ********* +-+
500 +-+-********************************-+-+
0 1 2 3 4 5
I have string variables MIN and SEC (minute and seconds).
MIN = "1"
SEC = "34"
I want to do calculations on these.
TOTSEC = MIN*60 + SEC
I tried:
expr $SEC + $MIN * 60
Result:
expr: non-numeric argument
Let it be known I am running busybox on a custom microcomputer and so have no access to bash,bc, and that other solution provides.
In sh, by which I'll assume you mean a POSIX shell, your best option is to use Arithmetic Expansion:
$ MIN=1
$ SEC=34
$ TOTSEC=$(( MIN * 60 + SEC ))
$ printf '%d\n' "$TOTSEC"
94
In csh however, the built-in math works quite differently:
% set MIN = 1
% set SEC = 34
% # TOTSEC = ( $MIN * 60 + $SEC )
% printf '%d\n' "$TOTSEC"
94
According to the man page, the # command permits numeric calculations to be performed and the result assigned to a variable.
Note that the expr command is external to the shell, so it should be usable from either one.
In sh:
$ TOTSEC=$(expr "$MIN" \* 60 + "$SEC")
And in csh:
% set TOTSEC = `expr "$MIN" \* 60 + "$SEC"`
Note: your sh may not be POSIX compliant. Most likely, it's ash, which is the ancestor of dash and FreeBSD's /bin/sh. You'll need to test in your environment.
I got a string as follow :
foo=0j0h0min0s
What would be the best way to convert it in seconds without using date ?
I tried something like this that sounded pretty nice but no luck :
#> IFS=: read -r j h min s <<<"$foo"
#> time_s=$((((j * 24 + h) * 60 + min) * 60 + s))
ksh: syntax error: `<' unexpected
Any idea is welcome, I just can't use date -d to make conversion as it is not present on the system I am working on.
<<<"$foo" is mainly a bash-ism. It is supported in some/newer ksh. (google 'ksh here string' ).
Your read is trying to split at :, wich is not present in your input
If you first get rid of characters, you can split at blank (as ususal)
and changing the here-string to a here-doc
#!/bin/ksh
foo=1j2h3min4s
read -r j h min s << END
"${foo//[a-z]/ }"
END
# or echo "${foo//[a-z]/ }" | read -r j h min s
time_s=$((((j * 24 + h) * 60 + min) * 60 + s))
echo ">$foo< = >${foo//[a-z]/ }< = $j|$h|$min|$s => >$time_s<"
>1j2h3min4s< = >1 2 3 4 < = "1|2|3|4 " => >93784<
# or using array, easy to assign, more typing where used
typeset -a t=( ${foo//[a-z]/ } )
time_s=$(( (( t[0] * 24 + t[1]) * 60 + t[2]) * 60 + t[3] ))
echo ">$foo< = >${foo//[a-z]/ }< = ${t[0]}|${t[1]}|${t[2]}|${t[3]} => >$time_s<"
I'm learning Shell scripting, I'm trying to write a small script that adds and multi number as shown below But amt value not display
value=212
amt=`expr "( $value * 2 + ( $value * 2 * .075 ) ) " | bc`
echo $amt
Your code works fine, but I suggest some improvements:
Use $(...) instead of backticks.
Replace expr with echo.
Example:
value=212
amt=$(echo "( $value * 2 + ( $value * 2 * .075 ) ) " | bc)
echo $amt
Output:
455.800
First things first. This:
value * 2 + (value * 2 * .075)
is a hot mess, this is equivalent:
value * 43 / 20
Next, I prefer to use awk for this job:
#!/usr/bin/awk -f
BEGIN {
value = 212
print value * 43 / 20
}
How do I truncate a floating point number using bc
e.g if I do
echo '4.2-1.3' | bc
which outputs 2.9 how I get it to truncate/use floor to get 2
Use / operator.
echo '(4.2-1.3) / 1' | bc
Dividing by 1 works ok if scale is 0 (eg, if you start bc with bc and don't change scale) but fails if scale is positive (eg, if you start bc with bc -l or increase scale). (See transcript below.) For a general solution, use a trunc function like the following:
define trunc(x) { auto s; s=scale; scale=0; x=x/1; scale=s; return x }
Transcript that illustrates how divide by 1 by itself fails in the bc -l case, but how trunc function works ok at truncating toward zero:
> bc -l
bc 1.06.95
[etc...]
for (x=-4; x<4; x+=l(2)) { print x,"\t",x/1,"\n"}
-4 -4.00000000000000000000
-3.30685281944005469059 -3.30685281944005469059
-2.61370563888010938118 -2.61370563888010938118
-1.92055845832016407177 -1.92055845832016407177
-1.22741127776021876236 -1.22741127776021876236
-.53426409720027345295 -.53426409720027345295
.15888308335967185646 .15888308335967185646
.85203026391961716587 .85203026391961716587
1.54517744447956247528 1.54517744447956247528
2.23832462503950778469 2.23832462503950778469
2.93147180559945309410 2.93147180559945309410
3.62461898615939840351 3.62461898615939840351
define trunc(x) { auto s; s=scale; scale=0; x=x/1; scale=s; return x }
for (x=-4; x<4; x+=l(2)) { print x,"\t",trunc(x),"\n"}
-4 -4
-3.30685281944005469059 -3
-2.61370563888010938118 -2
-1.92055845832016407177 -1
-1.22741127776021876236 -1
-.53426409720027345295 0
.15888308335967185646 0
.85203026391961716587 0
1.54517744447956247528 1
2.23832462503950778469 2
2.93147180559945309410 2
3.62461898615939840351 3
Try the following solution. It will truncate anything after the decimal point without a problem:
echo 'x = 4.2 - 1.3; scale = 0; x / 1' | bc -l
echo 'x = l(101) / l(10); scale = 0; x / 1' | bc -l
You can make the code a tad shorter by performing calculations directly on the numbers:
echo 'scale = 0; (4.2 - 1.3) / 1' | bc -l
echo 'scale = 0; (l(101) / l(10)) / 1' | bc -l
In general, you can use this function to get only the integer part of a number:
define int(x) {
auto s;
s = scale;
scale = 0;
x /= 1; /* This will have the effect of truncating x to its integer value */
scale = s;
return (x);
}
Save that code into a file (let's call it int.bc) and run the following command:
echo 'int(4.2 - 1.3);' | bc -l int.bc
The variable governing the amount of decimals on division is scale.
So, if scale is 0 (the default), dividing by 1 would truncate to 0 decimals:
$ echo '(4.2-1.3) / 1 ' | bc
2
In other operations, the number of decimals is calculated from the scale (number of decimals) of each operand. In add, subtract and multiplication, for example, the resulting scale is the biggest of both:
$ echo ' 4.2 - 1.33333333 ' | bc
2.86666667
$ echo ' 4.2 - 1.333333333333333333 ' | bc
2.866666666666666667
$ echo ' 4.2000 * 1.33 ' | bc
5.5860
Instead, in division, the number of decimals is strictly equal to th evalue of the variable scale:
$ echo 'scale=0;4/3;scale=3;4/3;scale=10;4/3' | bc
1
1.333
1.3333333333
As the value of scale has to be restored, it is better to define a function (GNU syntax):
$ echo ' define int(x){ os=scale;scale=0;x=x/1;scale=os;return(x) }
int( 4.2-1.3 )' | bc
2
Or in older POSIX language:
$ echo ' define i(x){
o=scale;scale=0;x=x/1;scale=o;return(x)
}
i( 4.2-1.3 )' | bc
2
You say:
truncate/use floor
And those are not the same thing in all cases. The other answers so far only show you how to truncate (i.e. "truncate towards zero" i.e. "discard the part after the decimal").
For negative numbers, the behavior is different.
To wit:
truncate(-2.5) = -2
floor(-2.5) = -3
So, here is a floor function for bc:
# Note: trunc(x) is defined as noted elsewhere in the other answers
define floor(x) {
auto t
t=trunc(x)
if (t>x) {
return t-1
} else {
return t
}
}
Aside:
You can put this, and other helper functions, in a file. For instance, I have this alias in my shell:
alias bc='bc -l ~/.bcinit'
And so whenever I run bc, I get all my utility functions from ~/.bcinit available by default.
Also, there is a good list of bc functions here: http://phodd.net/gnu-bc/code/funcs.bc
You may do something like this:
$ printf "%.2f\n" $(echo "(4530 / 4116 - 1) * 100" | bc -l)
10.06
Here I am trying to find the % change. Not purely bc though.