Sort each column independently - sorting

cat sanger.* | tr '\-ACGT' '01234' | sed -e 's/\([[:digit:]]\)/\1 /g'
1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 1 1 1 0
0 2 2 0 0 0 0 2 2 2 2 0 2 0 0 0 0 0 2 2 2 0 2 0 0 0 0 0 0 0 2
0 0 0 0 0 0 3 0 0 0 0 3 0 0 3 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0
0 0 0 4 4 0 0 0 0 0 0 0 0 4 0 4 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0
This is my current output, now I want to sort each column independently, so all the numbers are on the same line.
How can I do that?


I am not sorting, here, but extracting the non-0 digits.
Here is an awk filter that "updates" each fields with the only (actually, the latest) non-"0" content it sees :
# short version
awk '/./ { if ( NF > maxNF ) { maxNF=NF }
for(i=1;i<=NF;i++) { if ( $i!="0" ) { result[i]=$i } }
}
END { for(i=1;i<=maxNF;i++) { printf "%s ",result[i] } }
'
# expanded version (ie, the same as above, with different indentation to mhelp reading)
awk '/./ { if ( NF > maxNF )
{ maxNF=NF }
for(i=1;i<=NF;i++)
{ if ( $i!="0" )
{ result[i]=$i }
}
}
END { for(i=1;i<=maxNF;i++)
{ printf "%s ",result[i]
}
}
'
so if I paste your posted result into that filter:
echo "
1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 1 1 1 0
0 2 2 0 0 0 0 2 2 2 2 0 2 0 0 0 0 0 2 2 2 0 2 0 0 0 0 0 0 0 2
0 0 0 0 0 0 3 0 0 0 0 3 0 0 3 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0
0 0 0 4 4 0 0 0 0 0 0 0 0 4 0 4 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0
" | awk '/./ { if ( NF > maxNF ) { maxNF=NF }
for(i=1;i<=NF;i++) { if ( $i!="0" ) { result[i]=$i } }
}
END { for(i=1;i<=maxNF;i++) { printf "%s ",result[i] } }
'
it outputs:
1 2 2 4 4 1 3 2 2 2 2 3 2 4 3 4 1 3 2 2 2 1 2 1 3 4 1 1 1 1 2
(note: with an extra " " at the end, here...)
A note of warning however: very OLD version of the original awk (and maybe some nawk) are limited to 99 fields... (Rarely encountered nowadays. And if you use GNU's version, you will be fine)

Related

Algorithm to divide black and white chocolate bar with least number of breaks

I have a rectangular chocolate bar that consists of squares either black, white or mixed. The bar being not bigger than 50x50 squares. I'm supposed to divide the bar between two people(one gets all the white squares and one the black ones, mixed ones don't matter), by cracking it either horizontally or vertically. I'm supposed to find a method with the least amount of such cracks.
I'm given this input:
M N (number of rows, number of columns)
and then M rows that are N numbers long(0 means white, 1 means black, 2 is mixed)
so for example bar described like this:
4 4
0 1 1 1
1 0 1 0
1 0 1 0
2 0 0 0
can be divided by cracking it seven times in total.
And this one needs at least 24 cracks:
5 8
0 1 0 1 0 1 0 2
1 0 2 0 2 0 1 0
0 2 0 2 0 1 0 2
1 0 2 0 2 0 1 0
0 1 0 1 0 1 0 2
I was thinking about something like cracking the bar in two pieces so that the sum of sums of future cracks needed to divide two newly made chocolate bar pieces is the least possible from all the possible cracks (which is height -1 + width -1 of the current bar piece being cracked)
I managed, also thanks to zenwraight, to write a code that solves this but I encountered another problem, it is really inefficient and if the starting chocolate bar gets bigger than 30x30 it's practically unusable.
Anyway the source code (written in C):
#include <stdio.h>
#include <stdlib.h>
const int M, N;
int ****pieces;
int r = 0;
int ri = 0;
int inf;
void printmatrix(int **mat, int starti, int startj, int maxi, int maxj) {
for (int i = starti; i < maxi; i++) {
for (int j = startj; j < maxj; j++) {
printf("%d ", mat[i][j]);
}
printf("\n");
}
}
int minbreaks(int **mat, int starti, int startj, int maxi, int maxj, int depth) {
if (pieces[starti][startj][maxi][maxj] != 0) {
r++;
return pieces[starti][startj][maxi][maxj];
} else {
ri++;
int vbreaks[maxj - 1];
int hbreaks[maxi - 1];
for (int i = 0; i < maxj; i++) {
vbreaks[i] = inf;
}
for (int i = 0; i < maxi; i++) {
hbreaks[i] = inf;
}
int currentmin = inf;
for (int i = starti; i < maxi; i++) {
for (int j = startj; j < maxj - 1; j++) {//traverse trough whole matrix
if (mat[i][j] != 2) {
for (int k = startj + 1; k < maxj; k++) {//traverse all columns
if (vbreaks[k - 1] == inf) {//traverse whole column
for (int z = starti; z < maxi; z++) {
if (mat[z][k] != 2 && mat[i][j] != mat[z][k]) {
/* printmatrix(mat, starti, startj, maxi, maxj);
printf("brokenv in depth:%d->\n", depth);
printmatrix(mat, starti, startj, maxi, k);
printf("and\n");
printmatrix(mat, starti, k, maxi, maxj);
printf("****\n");*/
vbreaks[k - 1] = minbreaks(mat, starti, startj, maxi, k, depth + 1) + minbreaks(mat, starti, k, maxi, maxj, depth + 1);
if (vbreaks[k - 1] < currentmin) {
currentmin = vbreaks[k - 1];
}
break;
}
}
}
}
}
}
}
for (int i = starti; i < maxi - 1; i++) {
for (int j = startj; j < maxj; j++) {
if (mat[i][j] != 2) {
for (int k = starti + 1; k < maxi; k++) {
if (hbreaks[k - 1] == inf) {
for (int z = startj; z < maxj; z++) {
if (mat[k][z] != 2 && mat[i][j] != mat[k][z]) {
/* printmatrix(mat, starti, startj, maxi, maxj);
printf("brokenh in depth:%d->\n", depth);
printmatrix(mat, starti, startj, k, maxj);
printf("and\n");
printmatrix(mat, k, startj, maxi, maxj);
printf("****\n");*/
hbreaks[k - 1] = minbreaks(mat, starti, startj, k, maxj, depth + 1) + minbreaks(mat, k, startj, maxi, maxj, depth + 1);
if (hbreaks[k - 1] < currentmin) {
currentmin = hbreaks[k - 1];
}
break;
}
}
}
}
}
}
}
if (currentmin == inf) {
currentmin = 1;
}
pieces[starti][startj][maxi][maxj] = currentmin;
return currentmin;
}
}
void alloc(int i, int j) {
pieces[i][j] = malloc(sizeof (int*)*(M + 1));
for (int y = i; y < M + 1; y++) {
pieces[i][j][y] = malloc(sizeof (int)*(N + 1));
for (int x = j; x < N + 1; x++) {
pieces[i][j][y][x] = 0;
}
}
}
int main(void) {
FILE *file = fopen("pub08.in", "r");
//FILE *file = stdin;
fscanf(file, "%d %d", &M, &N);
int **mat = malloc(sizeof (int*)*M);
pieces = malloc(sizeof (int***)*M);
for (int i = 0; i < M; i++) {
mat[i] = malloc(sizeof (int)*N);
pieces[i] = malloc(sizeof (int**)*N);
for (int j = 0; j < N; j++) {
int x;
fscanf(file, "%d", &x);
mat[i][j] = x;
alloc(i, j);
}
}
inf = M * (M + 1) * N * (N + 1) / 4 + 1;
int result = minbreaks(mat, 0, 0, M, N, 0);
printf("%d\n", result);
printf("ri:%d,r:%d\n", ri, r);
return (EXIT_SUCCESS);
}
I am aiming to solve this input :
40 40
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 1 2 1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 1 1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 2 1 2 1 2 0 0 1 2 2 0 0 0 0 0 0 0 0 1 1 2 1 2 0 0 0 0 0 0 0 0 0 0
0 0 0 1 2 2 0 1 1 1 1 1 0 0 1 2 2 0 0 0 0 0 1 0 0 2 2 1 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 2 2 1 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1 1 2 2 0 0 0 1 2 2 1 2 1 0 0 0 0 0 1 2 1 2 0 0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 2 2 1 2 0 0 0 0 0 2 1 2 2 0 0 0 0 0 2 1 2 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 2 2 2 1 1 0 0 0 0 0 2 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0
0 2 1 2 1 0 2 2 2 2 1 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 1 2 0 2 2 1 0 0 0 0 0 0
0 2 2 1 2 0 1 2 2 1 1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 1 1 1 0 0 0 0 0 0
0 2 2 1 2 0 0 0 0 2 1 2 1 2 1 1 2 0 2 0 0 0 0 0 0 0 1 2 2 2 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 2 2 2 2 1 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 1 2 1 1 2 2 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 2 0 0 0 0
0 0 0 0 0 0 0 2 1 2 0 0 2 2 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 1 1 0 0 0 0
0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 2 2 0 0 0 0
0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 2 2 1 0 0 0 0 2 0 1 1 1 2 1 2 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 2 0 0 0 0 0 0 2 1 2 2 2 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 2 0 0 0 0 0 0 1 2 1 1 2 2 0 0 0 0 0
0 0 0 0 0 0 1 2 1 2 2 1 0 0 0 0 0 0 0 1 2 1 2 0 0 0 0 0 0 0 0 0 2 1 2 0 0 0 0 0
0 0 0 0 0 0 1 2 2 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 2 0 0 0 0 0 0 0 0 0 0 2 1 1 0 0
0 0 0 0 0 0 1 1 1 1 1 2 2 0 0 0 0 0 0 0 0 1 1 1 2 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0
0 0 0 0 0 0 1 2 2 2 1 1 1 0 0 0 0 0 0 0 0 1 2 1 2 0 0 0 0 0 0 0 0 0 0 2 2 2 1 0
0 0 0 0 0 0 0 0 0 1 2 1 2 0 0 0 0 0 0 0 0 1 1 1 2 2 0 0 0 0 0 0 0 0 0 1 2 1 1 0
0 0 0 2 1 1 2 2 0 1 2 1 1 0 0 0 0 0 2 2 1 2 2 1 2 2 0 0 0 0 0 0 0 0 0 1 2 2 2 0
0 0 0 2 2 2 1 1 0 0 1 2 2 2 0 0 0 0 2 2 2 1 1 2 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 2 1 2 2 1 1 0 2 1 2 1 2 1 2 1 1 2 1 1 1 1 1 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2 1 0 1 1 1 1 1 1 2 1 1 2 2 1 0 1 2 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 1 0 0 2 1 1 1 2 1 2 0 0 1 2 1 2 1 2 2 0 0 0 0 0 0 0 1 1 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 1 2 2 1 1 2 2 1 1 1 1 1 1 1 2 1 0 0 0 0 0 0 0 2 2 2 0 0 0
0 0 0 0 0 0 0 1 1 1 2 0 0 1 1 1 2 2 1 2 2 2 1 0 0 0 1 1 1 0 0 0 0 0 1 2 1 0 0 0
0 0 0 0 0 0 0 2 1 1 2 0 0 0 0 0 0 2 2 2 1 1 1 0 0 0 1 2 2 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 0 0 2 1 1 1 2 0 0 0 0 1 2 2 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 2 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 1 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 1 1 2 0 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 0 0 0 1 2 1 0 0
0 0 0 0 0 0 0 0 0 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 0 0 0 0 0 0 0 1 2 1 0 0
0 0 0 0 0 0 0 0 0 2 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
in under 2 seconds, which is much much faster time than that of my current program. The minimum amount of cracks for this one is 126.

Nice question, I have an approach in mind which makes use of recursion to tackle the above problem.
So as at each level or step you have two options either to split the bar horizontally or vertically.
So let's understand the algorithm with an example.
Example:-
4 4
0 1 1 1
1 0 1 0
1 0 1 0
2 0 0 0
Now let's call our function minBreaks(int n, int m, int matleft, int right)
So at first step if we break is horizontally our matleft will be
0
1
1
2
and matright will be
1 1 1
0 1 0
0 1 0
0 0 0
Now similarly if we had broken this vertically our matleft will be
0 1 1 1
and matright will be
1 0 1 0
1 0 1 0
2 0 0 0
Now you pass along this matleft and matright in next recursion call
And then at each call when size of row = 1 or col = 1, you can check the connected components of same value and return the count of connected components
Like for example for vertically case of maxleft -> 0 1 1 1, you will return 2.
Similarly for all the cases and the end part of the method will return
min between break horizontally and vertically
Hope this helps!

how to record properties of other variables in stata

I have to generate variables entry_1, entry_2 and entry_3 which will adopt the value 1 if id_i for that particular month had entry=1.
Example.
id month entry entry_1 entry_2 entry_3
1 1 1 1 0 0
1 2 0 0 0 0
1 3 0 0 1 1
1 4 0 0 0 0
2 1 0 1 0 0
2 2 0 0 0 0
2 3 1 0 1 1
2 4 0 0 0 0
3 1 0 1 0 0
3 2 0 0 0 0
3 3 1 0 1 1
3 4 0 0 0 0
Would anyone be so kind to propose an idea of how to implement a loop in order to do this?
I am thinking of something like this:
forvalues i=1(1)3 {
gen entry`i'=0
replace entry`i'=1 if on that particular month id=`i' had entry=1
}

You could do something like this (although your data don't quite look right for the question you're asking):
forvalues i = 1/3 {
gen entry_`i' = id == `i' & entry == 1
}
This generates a dummy variable entry_i for each i in the forvalues loop where entry_i = 1 if id is i and entry is 1, and 0 otherwise.

The code can be simplified down to at most one loop.
clear
input id month entry entry_1 entry_2 entry_3
1 1 1 1 0 0
1 2 0 0 0 0
1 3 0 0 1 1
1 4 0 0 0 0
2 1 0 1 0 0
2 2 0 0 0 0
2 3 1 0 1 1
2 4 0 0 0 0
3 1 0 1 0 0
3 2 0 0 0 0
3 3 1 0 1 1
3 4 0 0 0 0
end
forval j = 1/4 {
egen entry`j' = total(entry & id == `j'), by(month)
}
list id month entry entry? , sepby(id)
+--------------------------------------------------------+
| id month entry entry1 entry2 entry3 entry4 |
|--------------------------------------------------------|
1. | 1 1 1 1 0 0 0 |
2. | 1 2 0 0 0 0 0 |
3. | 1 3 0 0 1 1 0 |
4. | 1 4 0 0 0 0 0 |
|--------------------------------------------------------|
5. | 2 1 0 1 0 0 0 |
6. | 2 2 0 0 0 0 0 |
7. | 2 3 1 0 1 1 0 |
8. | 2 4 0 0 0 0 0 |
|--------------------------------------------------------|
9. | 3 1 0 1 0 0 0 |
10. | 3 2 0 0 0 0 0 |
11. | 3 3 1 0 1 1 0 |
12. | 3 4 0 0 0 0 0 |
+--------------------------------------------------------+

Bash: Pipe output into a table

I have a program that prints out the following:
bash-3.2$ ./drawgrid
0
1 1 0
1 1 0
0 0 0
1
0 1 1
0 1 1
0 0 0
2
0 0 0
1 1 0
1 1 0
3
0 0 0
0 1 1
0 1 1
Is it possible to pipe the output of this command such that I get all the 3x3 matrices (together with their number) displayed on a table, for example a 2x2 like this?
0 1
1 1 0 0 1 1
1 1 0 0 1 1
0 0 0 0 0 0
2 3
0 0 0 0 0 0
1 1 0 0 1 1
1 1 0 0 1 1
I tried searching, and came across the column command, but I did not figure it out.
Thank you

You can use pr -2T to get the following output, which is close to what you expected:
0 2
1 1 0 0 0 0
1 1 0 1 1 0
0 0 0 1 1 0
1 3
0 1 1 0 0 0
0 1 1 0 1 1
0 0 0 0 1 1

You could use an awk script:
NF == 1 {
if ($NF % 2 == 0) {
delete line
line[1]=$1
f=1
} else {
print line[1]"\t"$1
f=0
}
n=1
}
NF > 1 {
n++
if (f)
line[n]=$0
else
print line[n]"\t"$0
}
And pipe to it like so:
$ ./drawgrid | awk -f 2x2.awk
0 1
1 1 0 0 1 1
1 1 0 0 1 1
0 0 0 0 0 0
2 3
0 0 0 0 0 0
1 1 0 0 1 1
1 1 0 0 1 1

You can get exactly what you expect with a short bash script and a little array index thought:
#!/bin/bash
declare -a idx
declare -a acont
declare -i cnt=0
declare -i offset=0
while IFS=$'\n'; read -r line ; do
[ ${#line} -eq 1 ] && { idx+=( $line ); ((cnt++)); }
[ ${#line} -gt 1 ] && { acont+=( $line );((cnt++)); }
done
for ((i = 0; i < ${#idx[#]}; i+=2)); do
printf "%4s%8s\n" ${idx[i]} ${idx[i+1]}
for ((j = offset; j < offset + 3; j++)); do
printf " %8s%8s\n" ${acont[j]} ${acont[j+3]}
done
offset=$((j + 3))
done
exit 0
Output
$ bash array_cols.sh <dat/cols.txt
0 1
1 1 0 0 1 1
1 1 0 0 1 1
0 0 0 0 0 0
2 3
0 0 0 0 0 0
1 1 0 0 1 1
1 1 0 0 1 1

Use an awk loop to subset a file

I have a file with lots of pieces of information that I want to split on the first column.
Example (example.gen):
1 rs3094315 752566 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
1 rs2094315 752999 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
2 rs3044315 759996 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
2 rs3054375 799966 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
2 rs3094375 999566 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
3 rs3078315 799866 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
3 rs4054315 759986 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs4900215 752998 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs5094315 759886 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs6094315 798866 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
Desired output:
Chr1.gen
1 rs3094315 752566 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
1 rs2094315 752999 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
Chr2.gen
2 rs3044315 759996 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
2 rs3054375 799966 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
2 rs3094375 999566 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
Chr3.gen
3 rs3078315 799866 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
3 rs4054315 759986 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
Chr4.gen
4 rs4900215 752998 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs5094315 759886 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs6094315 798866 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
I've tried to do this with the following shell scripts, but it doesn't work - I can't work out how to get awk to recognise a variable defined outside the awk script itself.
First script attempt (no awk loop):
for i in {1..23}
do
awk '{$1 = $i}' example.gen > Chr$i.gen
done
Second script attempt (with awk loop):
for i in {1..23}
do
awk '{for (i = 1; i <= 23; i++) $1 = $i}' example.gen > Chr$i.gen
done
I'm sure its probably quite basic, but I just can't work it out...
Thank you!

With awk:
awk '{print > "Chr"$1".gen"}' file
It just prints and redirects it to a file. And how is this file defined? With "Chr" + first_column + ".gen".
With your sample input it creates 4 files. For example the 4th is:
$ cat Chr4.gen
4 rs4900215 752998 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs5094315 759886 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1
4 rs6094315 798866 A G 0 1 0 1 0 0 1 0 0 0 1 0 0 1

First, use #fedorqui's answer, as that is best. But to understand the mistake you made with your first attempt (which was close), read on.
Your first attempt failed because you put the test inside the action (in the braces), not preceding it. The minimal fix:
awk "\$1 == $i" example.gen > Chr$i.gen
This uses double quotes to allow the value of i to be seen by the awk script, but that requires you to then escape the dollar sign for $1 so that you don't substitute the value of the shell's first positional argument. Cleaner but longer:
awk -v i=$i '$1 == i' example.gen > Chr$i.gen
This adds creates a variable i inside the awk script with the same value as the shell's i variable.

how to convert these 7-segment decoder to boolean expression

how to convert these 7-segment decoder to boolean expression??
BCD 7-Segment decoder
A B C D a b c d e f g
0 0 0 0 0 0 0 0 0 0 1
0 0 0 1 1 0 0 1 1 1 1
0 0 1 0 0 0 1 0 0 1 0
0 0 1 1 0 0 0 0 1 1 0
0 1 0 0 1 0 0 1 1 0 0
0 1 0 1 0 1 0 0 1 0 0
0 1 1 0 0 1 0 0 0 0 0
0 1 1 1 0 0 0 1 1 1 1
1 0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 0 0 0 1 0 0

I suggest you use a karnough map.
You'll need to use one for each result column, so 7 4x4 tables.
There are even a few karough map generators on the web that you can use.

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