I have been trying to convert base 10 to base 2 and finding the base 10 integer that denotes most consecutive 1s
if __name__ == '__main__':
n = int(input().strip())``
outcomes = 0
biggest = 0
while n > 0:
if n % 2 == 1:
outcomes += 1
if outcomes > biggest:
biggest = outcomes
else:
result = 0
print(biggest)
those are my codes please correct where i'm wrong
Related
I am trying to list down all the prime numbers up till a specific number e.g. 1000. The code gets slower as the number increase. I am pretty sure it is because of the for loop where (number -1) is checked by all the prime_factors. Need some advise how I can decrease the processing time of the code for larger numbers. Thanks
import time
t0 = time.time()
prime_list = [2]
number = 0
is_not_prime = 0
count = 0
while number < 1000:
print(number)
for i in range (2,number):
count = 0
if (number%i) == 0:
is_not_prime = 1
if is_not_prime == 1:
for j in range (0,len(prime_list)):
if(number-1)%prime_list[j] != 0:
count += 1
if count == len(prime_list):
prime_list.append(number-1)
is_not_prime = 0
count = 0
break
number += 1
print(prime_list)
t1 = time.time()
total = t1-t0
print(total)
Your solution, on top of being confusing, is very inefficient - O(n^3). Please, use the Sieve of Eratosthenes. Also, learn how to use booleans.
Something like this (not optimal, just a mock-up). Essentially, you start with a list of all numbers, 1-1000. Then, you remove ones that are the multiple of something.
amount = 1000
numbers = range(1, amount)
i = 1
while i < len(numbers):
n = i + 1
while n < len(numbers):
if numbers[n] % numbers[i] == 0:
numbers.pop(n)
else:
n += 1
i += 1
print(numbers)
Finally, I was able to answer because your question isn't language-specific, but please tag the question with the language you're using in the example.
Is there an equation I can use for arbitrary M and N?
Example, N=3 and M=2:
3 bits allow for 8 different combinations, but only 2 of them do not contain more than 2 same symbols in a row
000 - Fails
001 - Fails
010 - OK
011 - Fails
100 - Fails
101 - OK
110 - Fails
111 - Fails
One way to frame the problem is as follows: we would like to count binary words of length n without runs of length m or larger. Let g(n, m) denote the number of such words. In the example, n = 3 and m = 2.
If n < m, every binary word works, and we get g(n, m) = 2^n words in total.
When n >= m, we can choose to start with 1, 2, ... m-1 repeated values,
followed by g(n-1, m), g(n-2, m), ... g(n-m+1, m) choices respectively. Combined, we get the following recursion (in Python):
from functools import lru_cache
#lru_cache(None) # memoization
def g(n, m):
if n < m:
return 2 ** n
else:
return sum(g(n-j, m) for j in range(1, m))
To test for correctness, we can compute the number of such binary sequences directly:
from itertools import product, groupby
def brute_force(n, k):
# generate all binary sequences of length n
products = product([0,1], repeat=n)
count = 0
for prod in products:
has_run = False
# group consecutive digits
for _, gp in groupby(prod):
gp_size = sum(1 for _ in gp)
if gp_size >= k:
# there are k or more consecutive digits in a row
has_run = True
break
if not has_run:
count += 1
return count
assert 2 == g(3, 2) == brute_force(3, 2)
assert 927936 == g(20, 7) == brute_force(20, 7)
I have this function implemented in Cython:
def count_kmers_cython(str string, list alphabet, int kmin, int kmax):
"""
Count occurrence of kmers in a given string.
"""
counter = {}
cdef int i
cdef int j
cdef int N = len(string)
limits = range(kmin, kmax + 1)
for i in range(0, N - kmax + 1):
for j in limits:
kmer = string[i:i+j]
counter[kmer] = counter.get(kmer, 0) + 1
return counter
Can I do better with cython? Or Can I have any away to improve it?
I am new to cython, that is my first attempt.
I will use this to count kmers in DNA with alphabet restrict to 'ACGT'. The length of the general input string is the average bacterial genomes (130 kb to over 14 Mb, where each 1 kb = 1000 bp).
The size of the kmers will be 3 < kmer < 16.
I wish to know if I could go further and maybe use cython in this function to:
def compute_kmer_stats(kmer_list, counts, len_genome, max_e):
"""
This function computes the z_score to find under/over represented kmers
according to a cut off e-value.
Inputs:
kmer_list - a list of kmers
counts - a dictionary-type with k-mers as keys and counts as values.
len_genome - the total length of the sequence(s).
max_e - cut off e-values to report under/over represented kmers.
Outputs:
results - a list of lists as [k-mer, observed count, expected count, z-score, e-value]
"""
print(colored('Starting to compute the kmer statistics...\n',
'red',
attrs=['bold']))
results = []
# number of tests, used to convert p-value to e-value.
n = len(list(kmer_list))
for kmer in kmer_list:
k = len(kmer)
prefix, sufix, center = counts[kmer[:-1]], counts[kmer[1:]], counts[kmer[1:-1]]
# avoid zero division error
if center == 0:
expected = 0
else:
expected = (prefix * sufix) // center
observed = counts[kmer]
sigma = math.sqrt(expected * (1 - expected / (len_genome - k + 1)))
# avoid zero division error
if sigma == 0.0:
z_score = 0.0
else:
z_score = ((observed - expected) / sigma)
# pvalue for all kmers/palindromes under represented
p_value_under = (math.erfc(-z_score / math.sqrt(2)) / 2)
# pvalue for all kmers/palindromes over represented
p_value_over = (math.erfc(z_score / math.sqrt(2)) / 2)
# evalue for all kmers/palindromes under represented
e_value_under = (n * p_value_under)
# evalue for all kmers/palindromes over represented
e_value_over = (n * p_value_over)
if e_value_under <= max_e:
results.append([kmer, observed, expected, z_score, p_value_under, e_value_under])
elif e_value_over <= max_e:
results.append([kmer, observed, expected, z_score, p_value_over, e_value_over])
return results
OBS - Thank you CodeSurgeon by the help. I know there are other tools to count kmer efficiently but I am learning Python so I am trying to write my own functions and code.
I am trying to improve my programming skills by writing functions in multiple ways, this teaches me new ways of writing code but also understanding other people's style of writing code. Below is a function that calculates the sum of all even numbers in a fibonacci sequence up to the max value. Do you have any recommendations on writing this algorithm differently, maybe more compactly or more pythonic?
def calcFibonacciSumOfEvenOnly():
MAX_VALUE = 4000000
sumOfEven = 0
prev = 1
curr = 2
while curr <= MAX_VALUE:
if curr % 2 == 0:
sumOfEven += curr
temp = curr
curr += prev
prev = temp
return sumOfEven
I do not want to write this function recursively since I know it takes up a lot of memory even though it is quite simple to write.
You can use a generator to produce even numbers of a fibonacci sequence up to the given max value, and then obtain the sum of the generated numbers:
def even_fibs_up_to(m):
a, b = 0, 1
while a <= m:
if a % 2 == 0:
yield a
a, b = b, a + b
So that:
print(sum(even_fibs_up_to(50)))
would output: 44 (0 + 2 + 8 + 34 = 44)
I'm getting an error on line 3 "TypeError: 'int' object is not iterable," and its been bothering me. Any advice/fixes appreciated.
Example test: collatz_counts(4) → 3 # 4 -> 2 -> 1 (3 steps)
Code I have:
def collatz_counts(x):
num = 0
for i in (x):
if i == 1:
num += 1
return num
elif i % 2 == 0:
num(i) / 2
num += 1
num.append(i)
else:
num = (i*2) + 3
num += 1
num.append(i)
return num
This can be solved recursively:
def collatz_length(n):
if n == 1:
return 1
return 1 + collatz_length(3*n+1 if n%2 else n//2)
Which lends itself to be memoized if you are going to be calling for a range of numbers, e.g. in Py3
import functools as ft
#ft.lru_cache(maxsize=None)
def collatz_length(n):
if n == 1:
return 1
return 1 + collatz_length(3*n+1 if n%2 else n//2)
Which will run through the first million collatz sequences in about 2.31s vs about 28.6s for the iterative solution.
Use a while loop. Just modify x in place until you get to 1 and keep track of the number of steps each time you run a cycle.
def collatz_counts(x):
steps = 0
while x != 1:
if x % 2:
x = x * 3 + 1
else:
x = x // 2
steps += 1
return steps