Related
I'm wondering if anyone knows any tools which allow me to count the frequency of amino acids at any specific position in a multiple-sequence alignment.
For example if I had three sequences:
Species 1 - MMRSA
Species 2 - MMLSA
Species 3 - MMRTA
I'd like for a way to search by position for the following output:
Position 1 - M = 3;
Position 2 - M = 3;
Position 3 - R = 2, L = 1;
Position 4 - S = 2, T = 1;
Position 5 - A = 3.
Thanks! I'm familiar with R and Linux, but if there's any other software that can do this I'm sure I can learn.
Using R:
x <- read.table(text = "Species 1 - MMRSA
Species 2 - MMLSA
Species 3 - MMRTA")
ixCol = 1
table(sapply(strsplit(x$V4, ""), "[", ixCol))
# M
# 3
ixCol = 4
table(sapply(strsplit(x$V4, ""), "[", ixCol))
# S T
# 2 1
Depending input file format, there are likely a purpose built bioconductor packages/functions.
That is really easy to parse, you can use any language of choice.
Here is an example in Python using a dict and Counter to assemble the data in a simple object.
from collections import defaultdict, Counter
msa = '''
Species 1 - MMRSA
Species 2 - MMLSA
Species 3 - MMRTA
'''
r = defaultdict(list) #dictionary having the sequences index as key and the list of aa found at that index as value
for line in msa.split('\n'):
line = line.strip()
if line:
sequence = line.split(' ')[-1]
for i, aa in enumerate(list(sequence)):
r[i].append(aa)
count = {k:Counter(v) for k,v in r.items()}
print(count)
#{0: Counter({'M': 3}), 1: Counter({'M': 3}), 2: Counter({'R': 2, 'L': 1}), 3: Counter({'S': 2, 'T': 1}), 4: Counter({'A': 3})}
To print the output as you specified:
for k, v in count.items():
print(f'Position {k+1} :', end=' ') #add 1 to start counting from 1 instead of 0
for aa, c in v.items():
print(f'{aa} = {c};', end=' ')
print()
It prints:
Position 1 : M = 3;
Position 2 : M = 3;
Position 3 : R = 2; L = 1;
Position 4 : S = 2; T = 1;
Position 5 : A = 3;
Triangular numbers are numbers which is number of things when things can be arranged in triangular shape.
For Example, 1, 3, 6, 10, 15... are triangular numbers.
o o o o o o o o o o is shape of n=4 triangular number
what I have to do is A natural number N is given and I have to print
N expressed by sum of triangular numbers.
if N = 4
output should be
1 1 1 1
1 3
3 1
else if N = 6
output should be
1 1 1 1 1 1
1 1 1 3
1 1 3 1
1 3 1 1
3 1 1 1
3 3
6
I have searched few hours and couldn't find answers...
please help.
(I am not sure this might help, but I found that
If i say T(k) is Triangular number when n is k, then
T(k) = T(k-1) + T(k-3) + T(k-6) + .... + T(k-p) while (k-p) > 0
and p is triangular number )
Here's Code for k=-1(Read comments below)
#include <iostream>
#include <vector>
using namespace std;
long TriangleNumber(int index);
void PrintTriangles(int index);
vector<long> triangleNumList(450); //(450 power raised by 2 is about 200,000)
vector<long> storage(100001);
int main() {
int n, p;
for (int i = 0; i < 450; i++) {
triangleNumList[i] = i * (i + 1) / 2;
}
cin >> n >> p;
cout << TriangleNumber(n);
if (p == 1) {
//PrintTriangles();
}
return 0;
}
long TriangleNumber(int index) {
int iter = 1, out = 0;
if (index == 1 || index == 0) {
return 1;
}
else {
if (storage[index] != 0) {
return storage[index];
}
else {
while (triangleNumList[iter] <= index) {
storage[index] = ( storage[index] + TriangleNumber(index - triangleNumList[iter]) ) % 1000000;
iter++;
}
}
}
return storage[index];
}
void PrintTriangles(int index) {
// What Algorithm?
}
Here is some recursive Python 3.6 code that prints the sums of triangular numbers that total the inputted target. I prioritized simplicity of code in this version. You may want to add error-checking on the input value, counting the sums, storing the lists rather than just printing them, and wrapping the entire routine into a function. Setting up the list of triangular numbers could also be done in fewer lines of code.
Your code saved time but worsened memory usage by "memoizing" the triangular numbers (storing and reusing them rather than always calculating them when needed). You could do the same to the sum lists, if you like. It is also possible to make this more in the dynamic programming style: find the sum lists for n=1 then for n=2 etc. I'll leave all that to you.
""" Given a positive integer n, print all the ways n can be expressed as
the sum of triangular numbers.
"""
def print_sums_of_triangular_numbers(prefix, target):
"""Print sums totalling to target, each after printing the prefix."""
if target == 0:
print(*prefix)
return
for tri in triangle_num_list:
if tri > target:
return
print_sums_of_triangular_numbers(prefix + [tri], target - tri)
n = int(input('Value of n ? '))
# Set up list of triangular numbers not greater than n
triangle_num_list = []
index = 1
tri_sum = 1
while tri_sum <= n:
triangle_num_list.append(tri_sum)
index += 1
tri_sum += index
# Print the sums totalling to n
print_sums_of_triangular_numbers([], n)
Here are the printouts of two runs of this code:
Value of n ? 4
1 1 1 1
1 3
3 1
Value of n ? 6
1 1 1 1 1 1
1 1 1 3
1 1 3 1
1 3 1 1
3 1 1 1
3 3
6
I am trying to plot a 3D graph between 2 scalars and one matrix for each of its entries. On compiling it is giving me "Submatrix incorrectly defined" error on line 11. The code:
i_max= 3;
u = zeros(4,5);
a1 = 1;
a2 = 1;
a3 = 1;
b1 = 1;
hx = linspace(1D-6,1D6,13);
ht = linspace(1D-6,1D6,13);
for i = 1:i_max
for j = 2:4
u(i+1,j)=u(i,j)+(ht*(a1*u(i,j))+b1+(((a2*u(i,j+1))-(2*a2*u(i,j))+(a2*u(i,j-1)))*(hx^-2))+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx^-1)));
plot(ht,hx,u(i+1,j));
end
end
Full error message:
-->exec('C:\Users\deba123\Documents\assignments and lecture notes\Seventh Semester\UGP\Scilab\Simulation1_Plot.sce', -1)
+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx^-1)))
!--error 15
Submatrix incorrectly defined.
at line 11 of exec file called by :
emester\UGP\Scilab\Simulation1_Plot.sce', -1
Please help.
For a 3-dimensional figure, you need 2 argument vectors and a matrix for the function values. So I expanded u to a tensor.
At every operation in your code, I added the current dimension of the term. Now, a transparent handling of you calculation is given. For plotting you have to use the plot3d (single values) or surf (surface) command.
In a 3-dim plot, you want two map 2 vectors (hx,ht) with dim n and m to an scalar z. Therefore you reach a (nxm)-matrix with your results. Is this, what you want to do? Currently, you have 13 values for each u(i,j,:) - entry, but you want (13x13) for every figure. Maybe the eval3d-function can help you.
i_max= 3;
u = zeros(4,5,13);
a1 = 1;
a2 = 1;
a3 = 1;
b1 = 1;
hx = linspace(1D-6,1D6,13); // 1 x 13
ht = linspace(1D-6,1D6,13); // 1 x 13
for i = 1:i_max
for j = 2:4
u(i+1,j,:)= u(i,j)...
+ ht*(a1*u(i,j))*b1... // 1 x 13
+(((a2*u(i,j+1)) -(2*a2*u(i,j)) +(a2*u(i,j-1)))*(hx.^-2))... // 1 x 13
+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx.^-1)) ... // 1 x 13
+ hx*ones(13,1)*ht; // added to get non-zero values
z = squeeze( u(i+1,j, : ))'; // 1x13
// for a 3d-plot: (1x13, 1x13, 13x13)
figure()
plot3d(ht,hx, z'* z ,'*' ); //
end
end
I am writing some functionality for a visual node based CAD program that will not allow for me to loop so I need a workaround to enumerate a list of numbers. I am an architect with very little programming experience so any help would be great.
A have an array of numbers(numArray) coming in as such 0,1,2,3,4... (first column) I need to take those numbers and convert them into their counterpart for column 1,2,3,4 without using any loops or nested loops.
numArray 1 2 3 4
-----------
0 = 0|0|0|0
1 = 0|0|0|1
2 = 0|0|0|2
3 = 0|0|0|3
4 = 0|0|1|0
5 = 0|0|1|1
6 = 0|0|1|2
7 = 0|0|1|3
8 = 0|0|2|0
9 = 0|0|2|1
10= 0|0|2|2
12= 0|0|2|3
13= 0|0|3|0
14= 0|0|3|1
15= 0|0|3|2
16= 0|1|3|3
17= 0|1|0|0
18= 0|1|0|1
19= 0|1|0|2
20= 0|1|0|3
21= 0|1|1|0
22= 0|1|1|1
23= 0|1|1|2
24= 0|1|1|3
I have figured out column 4 by implementing the following:
int column4 = numArray % 4;
this works and creates the numbers as such 0,1,2,3,0,1,2,3.... this is great however I am not sure how to use the num array coming in to produce column 3 2 and 1. Again I have very little programming experience so any help would be great.
You're converting the input to base 4 notation, so this will do the job:
int input[] = {0,1,2,3,4,5,6,7,8,9,10,12,13,14,15,16,17,18,19,20,21,22,23,24};
for (int i = 0; i < sizeof(input)/sizeof(input[0]); i++)
{
int c1, c2, c3, c4;
c4 = input[i] % 4;
c3 = (input[i] / 4) % 4;
c2 = (input[i] / 16) % 4;
c1 = (input[i] / 64) % 4;
printf("%d = %d\t%d\t%d\t%d\n", input[i], c1, c2, c3, c4);
}
Given an X, what math is needed to find its Y, using this table?
x
y
0
1
1
0
2
6
3
5
4
4
5
3
6
2
This is a language agnostic problem. I can't just store the array, and do the lookup. The input will always be the finite set of 0 to 6. It won't be scaling later.
This:
y = (8 - x) % 7
This is how I arrived at that:
x 8-x (8-x)%7
----------------
0 8 1
1 7 0
2 6 6
3 5 5
4 4 4
5 3 3
6 2 2
int f(int x)
{
return x["I#Velcro"] & 7;
}
0.048611x^6 - 0.9625x^5 + 7.340278x^4 - 26.6875x^3 + (45 + 1/9)x^2 - 25.85x + 1
Sometimes the simple ways are best. ;)
It looks like:
y = (x * 6 + 1) % 7
I don't really like the % operator since it does division so:
y = (641921 >> (x*3)) & 7;
But then you said something about not using lookup tables so maybe this doesn't work for you :-)
Update:
Since you want to actually use this in real code and cryptic numbers are not nice, I can offer this more maintainable variant:
y = (0x2345601 >> (x*4)) & 15;
Though it seems a bunch of correct answers have already appeared, I figured I'd post this just to show another way to have worked it out (they're all basically variations on the same thing):
Well, the underlying pattern is pretty simple:
x y
0 6
1 5
2 4
3 3
4 2
5 1
6 0
y = 6 - x
Your data just happens to have the y values shifted "down" by two indices (or to have the x values shifted "up").
So you need a function to shift the x value. This should do it:
x = (x + 5) % 7;
Resulting equation:
y = 6 - ((x + 5) % 7);
Combining the ideas in Dave and Paul's answer gives the rather elegant:
y = (8 - x) % 7`
(though I see I was beaten to the punch with this)
unsigned short convertNumber(unsigned short input) {
if (input <= 1) { return !input; } //convert 0 => 1, 1 => 0
return (8-input); //convert 2 => 6 ... 6 => 2
}
Homework?
How about:
y = (x <= 1 ? 1 : 8) - x
and no, i dont/cant just store the array, and do the lookup.
Why not?
yes, the input will always be the finite set of 0 to 6. it wont be scaling later.
Just use a bunch of conditionals then.
if (input == 0) return 1;
else if (input == 1) return 0;
else if (input == 2) return 6;
...
Or find a formula if it's easy to see one, and it is here:
if (input == 0) return 1;
else if (input == 1) return 0;
else return 8 - input;
Here's a way to avoid both modulo and conditionals, going from this:
y = (8 - x) % 7
We know that x % y = x - floor(x/y)*y
So we can use y = 8 - x - floor((8 - x) / 7) * 7
What about some bit-fu ?
You can get the result using only minus, logical operators and shifts.
b = (x >> 2) | ((x >> 1) & 1)
y = ((b << 3)|(b ^ 1)) - x