Sorry if I don't write good, it's my first post.
I have a list in one file with the name, id, marks etc of students (see below):
And I want to calculate the average mark in another file, but I don't know how to take only the marks and write the average in another file.
Thanks;
#name surname student_index_number course_group_id lecturer_id list_of_marks
athos musketeer 1 1 1 3,4,5,3.5
porthos musketeer 2 1 1 2,5,3.5
aramis musketeer 3 2 2 2,1,4,5
while read line; do
echo "$line" | cut -f 6 -d ' '
done<main_list
awk 'NR>1{n=split($NF,a,",");for(i=1;i<=n;i++){s+=a[i]} ;print $1,s/n;s=0}' input
athos 3.875
porthos 3.5
aramis 3
For all the lines except header(NR>1 will filter out header) , pick up the last column and split into smaller numbers by comma. Using for loop sum the value of all the marks and then divid by the total subject number.
Something like (untested)
awk '{ n = split($6, a, ","); total=0; for (v in a) total += a[v]; print total / n }' main_list
In pure BASH solution, could you please try following once.
while read first second third fourth fifth sixth
do
if [[ "$first" =~ (^#) ]]
then
continue
fi
count="${sixth//[^,]}"
val=$(echo "(${#count}+1)" | bc)
echo "scale=2; (${sixth//,/+})/$val" | bc
done < "Input_file"
Related
I have a array=(4,2,8,9,1,0) and I don't want to sort the array to find the highest number in the array because I need to get the index value of the highest number as it is, so I can use it for further reference.
Expected output:
9 index value => 3
Can somebody help me to achieve this?
Slight variation with a loop using the ternary conditional operator and no assumptions about range of values:
arr=(4 2 8 9 1 0)
max=${arr[0]}
maxIdx=0
for ((i = 1; i < ${#arr[#]}; ++i)); do
maxIdx=$((arr[i] > max ? i : maxIdx))
max=$((arr[i] > max ? arr[i] : max))
done
printf '%s index => values %s\n' "$maxIdx" "$max"
The only assumption is that array indices are contiguous. If they aren't, it becomes a little more complex:
arr=([1]=4 [3]=2 [5]=8 [7]=9 [9]=1 [11]=0)
indices=("${!arr[#]}")
maxIdx=${indices[0]}
max=${arr[maxIdx]}
for i in "${indices[#]:1}"; do
((arr[i] <= max)) && continue
maxIdx=$i
max=${arr[i]}
done
printf '%s index => values %s\n' "$maxIdx" "$max"
This first gets the indices into a separate array and sets the initial maximum to the value corresponding to the first index; then, it iterates over the indices, skipping the first one (the :1 notation), checks if the current element is a new maximum, and if it is, stores the index and the maximum.
Without using sort, you can use a simple loop in shell. Here is a sample bash code:
#!/usr/bin/env bash
array=(4 2 8 9 1 0)
for i in "${!array[#]}"; do
[[ -z $max ]] || (( ${array[i]} > $max )) && { max="${array[i]}"; maxind=$i; }
done
echo "max=$max, maxind=$maxind"
max=9, maxind=3
arr=(4 2 8 9 1 0)
paste <(printf "%s\n" "${arr[#]}") <(seq 0 $((${#arr[#]} - 1)) ) |
sort -k1,1 |
tail -n1 |
sed 's/\t/ index value => /'
Print each array element on a newline with printf
Print array indexes with seq
Join both streams using paste
Numerically sort the lines using the first fields (ie. array value) sort
Print the last line tail -n1
The array value and result is separated by a tab. Substitute tab with the output string you want using sed. One could use ex. cut -d, -f2 to get only the index or use read a b <( ... ) to read the numbers into variables, etc.
Using Perl
$ export data=4,2,8,9,1,0
$ echo $data | perl -ne ' map{$i++; if($_>$x) {$x=$_;$id=$i} } split(","); print "max=$x", " index=",--${id},"\n" '
max=9 index=3
$
I have a big txt file with 2 columns and more than 2 million rows. Every value represents an id and there may be duplicates. There are about 100k unique ids.
1342342345345 34523453452343
0209239498238 29349203492342
2349234023443 99203900992344
2349234023443 182834349348
2923000444 9902342349234
I want to identify each id and re-number all of them starting from 1. It should re-number duplicates also using the same new id. If possible, it should be done using bash.
The output could be something like:
123 485934
34 44834
167 34564
167 2345
2 34564
Doing this in pure bash will be really slow. I'd recommend:
tr -s '[:blank:]' '\n' <file |
sort -un |
awk '
NR == FNR {id[$1] = FNR; next}
{for (i=1; i<=NF; i++) {$i = id[$i]}; print}
' - file
4 8
3 7
5 9
5 2
1 6
With bash and sort:
#!/bin/bash
shopt -s lastpipe
declare -A hash # declare associative array
index=1
# read file and fill associative array
while read -r a b; do
echo "$a"
echo "$b"
done <file | sort -nu | while read -r x; do
hash[$x]="$((index++))"
done
# read file and print values from associative array
while read -r a b; do
echo "${hash[$a]} ${hash[$b]}"
done < file
Output:
4 8
3 7
5 9
5 2
1 6
See: man bash and man sort
Pure Bash, with a single read of the file:
declare -A hash
index=1
while read -r a b; do
[[ ${hash[$a]} ]] || hash[$a]=$((index++)) # assign index only if not set already
[[ ${hash[$b]} ]] || hash[$b]=$((index++)) # assign index only if not set already
printf '%s %s\n' "${hash[$a]}" "${hash[$b]}"
done < file > file.indexed
Notes:
the index is assigned in the order read (not based on sorting)
we make a single pass through the file (not two as in other solutions)
Bash's read is slower than awk; however, if the same logic is implemented in Perl or Python, it will be much faster
this solution is more CPU bound because of the hash lookups
Output:
1 2
3 4
5 6
5 7
8 9
Just keep a monotonic counter and a table of seen numbers; when you see a new id, give it the value of the counter and increment:
awk '!a[$1]{a[$1]=++N} {$1=a[$1]} !a[$2]{a[$2]=++N} {$2=a[$2]} 1' input
awk 'NR==FNR { ids[$1] = ++c; next }
{ print ids[$1], ids[$2] }
' <( { cut -d' ' -f1 renum.in; cut -d' ' -f2 renum.in; } | sort -nu ) renum.in
join the two columns into one then sort the that into numerical order (-n), and make unique (-u), before using awk to use this sequence to generate an array of mappings between old to new ids.
Then for each line in input, swap ids and print.
EDIT
I read the question that this is supposed to be a duplicate of (this one). I don't agree. In that question the aim is to get the frequencies of individual numbers in the column. However if I apply that solution to my problem, I'm still left with my initial problem of grouping the frequencies of the numbers in a particular range into the final histogram. i.e. if that solution tells me that the frequency of 0.45 is 2 and 0.44 is 1 (for my input data), I'm still left with the problem of grouping those two frequencies into a total of 3 for the range 0.4-0.5.
END EDIT
QUESTION-
I have a long column of data with values between 0 and 1.
This will be of the type-
0.34
0.45
0.44
0.12
0.45
0.98
.
.
.
A long column of decimal values with repetitions allowed.
I'm trying to change it into a histogram sort of output such as (for the input shown above)-
0.0-0.1 0
0.1-0.2 1
0.2-0.3 0
0.3-0.4 1
0.4-0.5 3
0.5-0.6 0
0.6-0.7 0
0.7-0.8 0
0.8-0.9 0
0.9-1.0 1
Basically the first column has the lower and upper bounds of each range and the second column has the number of entries in that range.
I wrote it (badly) as-
for i in $(seq 0 0.1 0.9)
do
awk -v var=$i '{if ($1 > var && $1 < var+0.1 ) print $1}' input | wc -l;
done
Which basically does a wc -l of the entries it finds in each range.
Output formatting is not a part of the problem. If I simply get the frequencies corresponding to the different bins , that will be good enough. Also please note that the bin size should be a variable like in my proposed solution.
I already read this answer and want to avoid the loop. I'm sure there's a much much faster way in awk that bypasses the for loop. Can you help me out here?
Following the same algorithm of my previous answer, I wrote a script in awk which is extremely fast (look at the picture).
The script is the following:
#!/usr/bin/awk -f
BEGIN{
bin_width=0.1;
}
{
bin=int(($1-0.0001)/bin_width);
if( bin in hist){
hist[bin]+=1
}else{
hist[bin]=1
}
}
END{
for (h in hist)
printf " * > %2.2f -> %i \n", h*bin_width, hist[h]
}
The bin_width is the width of each channel. To use the script just copy it in a file, make it executable (with chmod +x <namefile>) and run it with ./<namefile> <name_of_data_file>.
For this specific problem, I would drop the last digit, then count occurrences of sorted data:
cut -b1-3 | sort | uniq -c
which gives, on the specified input set:
2 0.1
1 0.3
3 0.4
1 0.9
Output formatting can be done by piping through this awk command:
| awk 'BEGIN{r=0.0}
{while($2>r){printf "%1.1f-%1.1f %3d\n",r,r+0.1,0;r=r+.1}
printf "%1.1f-%1.1f %3d\n",$2,$2+0.1,$1}
END{while(r<0.9){printf "%1.1f-%1.1f %3d\n",r,r+0.1,0;r=r+.1}}'
The only loop you will find in this algorithm is around the line of the file.
This is an example on how to realize what you asked in bash. Probably bash is not the best language to do this since it is slow with math. I use bc, you can use awk if you prefer.
How the algorithm works
Imagine you have many bins: each bin correspond to an interval. Each bin will be characterized by a width (CHANNEL_DIM) and a position. The bins, all together, must be able to cover the entire interval where your data are casted. Doing the value of your number / bin_width you get the position of the bin. So you have just to add +1 to that bin. Here a much more detailed explanation.
#!/bin/bash
# This is the input: you can use $1 and $2 to read input as cmd line argument
FILE='bash_hist_test.dat'
CHANNEL_NUMBER=9 # They are actually 10: 0 is already a channel
# check the max and the min to define the dimension of the channels:
MAX=`sort -n $FILE | tail -n 1`
MIN=`sort -rn $FILE | tail -n 1`
# Define the channel width
CHANNEL_DIM_LONG=`echo "($MAX-$MIN)/($CHANNEL_NUMBER)" | bc -l`
CHANNEL_DIM=`printf '%2.2f' $CHANNEL_DIM_LONG `
# Probably printf is not the best function in this context because
#+the result could be system dependent.
# Determine the channel for a given number
# Usage: find_channel <number_to_histogram> <width_of_histogram_channel>
function find_channel(){
NUMBER=$1
CHANNEL_DIM=$2
# The channel is found dividing the value for the channel width and
#+rounding it.
RESULT_LONG=`echo $NUMBER/$CHANNEL_DIM | bc -l`
RESULT=`printf '%.0f' $RESULT_LONG`
echo $RESULT
}
# Read the file and do the computuation
while IFS='' read -r line || [[ -n "$line" ]]; do
CHANNEL=`find_channel $line $CHANNEL_DIM`
[[ -z HIST[$CHANNEL] ]] && HIST[$CHANNEL]=0
let HIST[$CHANNEL]+=1
done < $FILE
counter=0
for i in ${HIST[*]}; do
CHANNEL_START=`echo "$CHANNEL_DIM * $counter - .04" | bc -l`
CHANNEL_END=`echo " $CHANNEL_DIM * $counter + .05" | bc`
printf '%+2.1f : %2.1f => %i\n' $CHANNEL_START $CHANNEL_END $i
let counter+=1
done
Hope this helps. Comment if you have other questions.
I'm trying to write a bash script that calculates the average of numbers by rows and columns. An example of a text file that I'm reading in is:
1 2 3 4 5
4 6 7 8 0
There is an unknown number of rows and unknown number of columns. Currently, I'm just trying to sum each row with a while loop. The desired output is:
1 2 3 4 5 Sum = 15
4 6 7 8 0 Sum = 25
And so on and so forth with each row. Currently this is the code I have:
while read i
do
echo "num: $i"
(( sum=$sum+$i ))
echo "sum: $sum"
done < $2
To call the program it's stats -r test_file. "-r" indicates rows--I haven't started columns quite yet. My current code actually just takes the first number of each column and adds them together and then the rest of the numbers error out as a syntax error. It says the error comes from like 16, which is the (( sum=$sum+$i )) line but I honestly can't figure out what the problem is. I should tell you I'm extremely new to bash scripting and I have googled and searched high and low for the answer for this and can't find it. Any help is greatly appreciated.
You are reading the file line by line, and summing line is not an arithmetic operation. Try this:
while read i
do
sum=0
for num in $i
do
sum=$(($sum + $num))
done
echo "$i Sum: $sum"
done < $2
just split each number from every line using for loop. I hope this helps.
Another non bash way (con: OP asked for bash, pro: does not depend on bashisms, works with floats).
awk '{c=0;for(i=1;i<=NF;++i){c+=$i};print $0, "Sum:", c}'
Another way (not a pure bash):
while read line
do
sum=$(sed 's/[ ]\+/+/g' <<< "$line" | bc -q)
echo "$line Sum = $sum"
done < filename
Using the numsum -r util covers the row addition, but the output format needs a little glue, by inefficiently paste-ing a few utils:
paste "$2" \
<(yes "Sum =" | head -$(wc -l < "$2") ) \
<(numsum -r "$2")
Output:
1 2 3 4 5 Sum = 15
4 6 7 8 0 Sum = 25
Note -- to run the above line on a given file foo, first initialize $2 like so:
set -- "" foo
paste "$2" <(yes "Sum =" | head -$(wc -l < "$2") ) <(numsum -r "$2")
I have a file with a bunch of paths that look like so:
7 /usr/file1564
7 /usr/file2212
6 /usr/file3542
I am trying to use sort to pull out and print the path(s) with the most occurrences. Here it what I have so far:
cat temp| sort | uniq -c | sort -rk1 > temp
I am unsure how to only print the highest occurrences. I also want my output to be printed like this:
7 1564
7 2212
7 being the total number of occurrences and the other numbers being the file numbers at the end of the name. I am rather new to bash scripting so any help would be greatly appreciated!
To emit only the first line of output (with the highest number, since you're doing a reverse numeric sort immediately prior), pipe through head -n1.
To remove all content which is not either a number or whitespace, pipe through tr -cd '0-9[:space:]'.
To filter for only the values with the highest number, allowing there to be more than one:
{
read firstnum name && printf '%s\t%s\n' "$firstnum" "$name"
while read -r num name; do
[[ $num = $firstnum ]] || break
printf '%s\t%s\n' "$num" "$name"
done
} < temp
If you want to avoid sort and you are allowed to use awk, then you can do this:
awk '{
if($1>maxcnt) {s=$1" "substr($2,10,4); maxcnt=$1} else
if($1==maxcnt) {s=s "\n"$1" "substr($2,10,4)}} END{print s}' \
temp