Bioinformatics

리눅스 OS가 설치된 서버에서 시스템 효율을 높이는 방법 중 하나로 캐시 메모리를 비우는 것이 있다.

리눅스 서버의 캐시 메모리를 비우는 방법은 명령어 기반으로 터미널에서 간단히 사용할 수 있다.

일단 메모리 사용량 확인하는 명령어는 다음과 같다.

$ free -m

캐시 메모리를 비우는 명령어는 다음과 같다.

$ sudo sh -c "echo 3 > /proc/sys/vm/drop_caches"

참조:

[리눅스 서버의 메모리 간단 관리 방법(사용량 확인/캐시삭제/정기관리)]

http://osasf.net/discussion/587/%EB%A6%AC%EB%88%85%EC%8A%A4-%EC%84%9C%EB%B2%84%EC%9D%98-%EB%A9%94%EB%AA%A8%EB%A6%AC-%EA%B0%84%EB%8B%A8-%EA%B4%80%EB%A6%AC-%EB%B0%A9%EB%B2%95-%EC%82%AC%EC%9A%A9%EB%9F%89-%ED%99%95%EC%9D%B8-%EC%BA%90%EC%8B%9C%EC%82%AD%EC%A0%9C-%EC%A0%95%EA%B8%B0%EA%B4%80%EB%A6%AC

["echo 3 > /proc/sys/vm/drop_caches" - Permission denied as root]

https://unix.stackexchange.com/questions/109496/echo-3-proc-sys-vm-drop-caches-permission-denied-as-root

Problem

A common substring of a collection of strings is a substring of every member of the collection. We say that a common substring is a longest common substring if there does not exist a longer common substring. For example, "CG" is a common substring of "ACGTACGT" and "AACCGTATA", but it is not as long as possible; in this case, "CGTA" is a longest common substring of "ACGTACGT" and "AACCGTATA".

Note that the longest common substring is not necessarily unique; for a simple example, "AA" and "CC" are both longest common substrings of "AACC" and "CCAA".

Given: A collection of k$k$ (k≤100$k\le 100$) DNA strings of length at most 1 kbp each in FASTA format.

Return: A longest common substring of the collection. (If multiple solutions exist, you may return any single solution.)

Sample Dataset

>Rosalind_1
GATTACA
>Rosalind_2
TAGACCA
>Rosalind_3
ATACA

Sample Output

AC

Problem explanation

Source

import time
start_time = time.time()
def FindCommonString(s1, s2):
    if len(s2) > len(s1):
        s1, s2 = s2, s1
    n = len(s2)
    for i in range(n):
        for j in range(i + 1):
            token = s2[j: n - i + j]
            if token in s1:
                return token
def reform(lines):
    for line in lines:
        if line.startswith(">"):
            a = line.replace(">", "")
            a = a.replace("\n", "")
            id.append(a)
        else:
            if len(id) > len(seq):
                line = line.replace("\n","")
                seq.append(line)
            else:
                line = line.replace("\n", "")
                seq[len(id) - 1] = seq[len(id) - 1] + line
    return id , seq

f = open("D:/gs/rosalind/rosalind_lcsm.txt","r")
lines = f.readlines()
id = []
seq = []
id, seq = reform(lines)
print(seq[1])
com = ""
for i in range(len(seq) -1):
    if com == "":
        com = FindCommonString(seq[i], seq[i+1])
        print(com)
    else:
        if len(com) >= len(FindCommonString(seq[i+1], com)):
            com = FindCommonString(seq[i+1],com)
print(com)
print("--- %s seconds ---" %(time.time() - start_time))
f.close()

Searching Through the Haystack

Problem

For a random variable X$X$ taking integer values between 1 and n$n$, the expected value of X$X$ is E(X)=∑nk=1k×Pr(X=k)$\mathrm{E}(X)=\sum _{k=1}^{n}k\times \mathrm{P}\mathrm{r}(X=k)$. The expected value offers us a way of taking the long-term average of a random variable over a large number of trials.

As a motivating example, let X$X$ be the number on a six-sided die. Over a large number of rolls, we should expect to obtain an average of 3.5 on the die (even though it's not possible to roll a 3.5). The formula for expected value confirms that E(X)=∑6k=1k×Pr(X=k)=3.5$\mathrm{E}(X)=\sum _{k=1}^{6}k\times \mathrm{P}\mathrm{r}(X=k)=3.5$.

More generally, a random variable for which every one of a number of equally spaced outcomes has the same probability is called a uniform random variable (in the die example, this "equal spacing" is equal to 1). We can generalize our die example to find that if X$X$ is a uniform random variable with minimum possible value a$a$ and maximum possible value b$b$, then E(X)=a+b2$\mathrm{E}(X)=\frac{a+b}{2}$. You may also wish to verify that for the dice example, if Y$Y$ is the random variable associated with the outcome of a second die roll, then E(X+Y)=7$\mathrm{E}(X+Y)=7$.

Given: Six nonnegative integers, each of which does not exceed 20,000. The integers correspond to the number of couples in a population possessing each genotype pairing for a given factor. In order, the six given integers represent the number of couples having the following genotypes:

1. AA-AA
2. AA-Aa
3. AA-aa
4. Aa-Aa
5. Aa-aa
6. aa-aa

Return: The expected number of offspring displaying the dominant phenotype in the next generation, under the assumption that every couple has exactly two offspring.

Sample Dataset

1 0 0 1 0 1

Sample Output

3.5

Problem explanation

couple = [18855, 19614, 16897, 18945, 16056, 16489]
ng = []
for i in range(len(couple)):
    if i < 3:
       ng.append(couple[i] * 2)
    elif i == 3:
        ng.append(couple[i] * 3 / 4 * 2)
    elif i == 4:
        ng.append(couple[i] / 2 * 2)
    else :
        ng.append(0)

print(sum(ng))

The Need for Averages