Heavy python beginner here. I want to create a simple function for a PIN guessing game that receives two 4-digit lists ( [guess], [answer] ) and returns a string with 4 letters stating how close I am to guessing the correct [answer] sequence (eg. Higher, True, Lower, Higher)
However, I get a new list for each string:
def checkNumbers(guess,right):
for n in range(4):
result = []
if guess[n] == right[n]:
result.append("T") #true
elif guess[n] < right[n]:
result.append("H") #higher
elif guess[n] > right[n]:
result.append("L") #lower
else:
result.append("F") #false
print (result)
return
checkNumbers([1,2,3,5],[2,2,1,6])
The result should look like this:
checkNumbers([1,2,3,4], [2, 2, 1 , 6]) #call function with ([guess], [answer])
'HTLH' #returns a string stating how accurate [guess] is to [answer] list
Result looks like this however:
checkNumbers([1,2,3,5],[2,2,1,6])
['H']
['T']
['L']
['H']
Thanks very much in advance for any help I could get.
you can use string instead of list or "".join()
def checkNumbers(guess, right):
result = ""
for n in range(4):
if guess[n] == right[n]:
result += "T" # true
elif guess[n] < right[n]:
result += "H" # higher
elif guess[n] > right[n]:
result += "L" # lower
else:
result += "F" # false
print(result)
but... maybe you want to use zip function
def checkNumbers(guess, right):
result = ""
for g, r in zip(guess, right):
if g == r:
result += "T" # true
elif g < r:
result += "H" # higher
elif g > r:
result += "L" # lower
else:
result += "F" # false
print(result)
Funny bonus here:
def checkNumbers(guess, right):
print("".join("THL"[(g > r) + (g != r)] for g, r in zip(guess, right)))
I don't get why you need else part...
Initiate the list and print the result outside of the loop:
def checkNumbers(guess, right):
result = []
for n in range(4):
# do loopy stuff
print (result)
return # not strictly necessary
If you do it inside, you are creating a new list on every iteration.
Related
I am running my code through an example set: [9,20,2,3]
I have been using the modulo operator (%) and my code is:
if index < len(the_list):
for m in the_list:
print(m)
if m % 3 == 0:
a = True
else:
a = False
return a
else:
return False
Args:
the_list (list): The input list
index (int): The index
Returns:
_type_: True: if the value is divisible by 3,
False: if the value is not divisible by 3,
False: if the index is invalid
One additional condition I must include is if the index position is larger than the length of the list to return False, but I believe I already have that properly integrated.
You can just use a list comprehension.
[(num%3 == 0) for num in myList]
here is a function that could help you :
def divisible_by_three(numbers):
for number in numbers:
if number % 3 == 0:
return True
else:
return False
This should work, just replace myList with your actual list
myList = [9,20,2,3]
finalList = []
for x in myList:
if x % 3 == 0:
finalList.append(True)
else:
finalList.append(False)
print(finalList)
I'm trying to create a function. Function; it will simply be designed to increase the last letter sequence from its position in the alphabet or letter list.
import time
def CountDown(text,reply=3):
abc = list("ABCDEFGHIJ")
c = 1
text_list = list(text)
while 1:
Index = abc.index(text_list[-c])
if not list(filter(lambda a: a!=abc[-1], text_list)):
return "".join(text_list)
if text_list[-c] == abc[-1]:
text_list[-c] = abc[0]
c += 1
continue
else:
s=1
while 1:
text_list[-c] = abc[(Index+s) if (Index+s)<len(abc) else 0]
if text_list.count(abc[(Index+s) if (Index+s)<len(abc) else 0])+1<reply:
break
s+=1
text_list[-c] = abc[(Index+s) if (Index+s)<len(abc) else 0]
return "".join(text_list)
if __name__ == "__main__":
code="ABHD"
while 1:
code=CountDown(code)
time.sleep(0.5)
print(code)
OUTPUT:
ABHE
ABHF
ABHG
ABHI
ABHJ
ABIA
ABIC
ABID
ABIE
ABIF
ABIG
ABIH
ABIJ
ABJA
ABJC
ABJD
ABJE
ABJF
ABJG
ABJH
ABJI
....(idling)
The code doesn't give an output after a while. I think there is something wrong.
How can I fix this code sample?
I have two strings
e.g.
str1 = "Come"
str2 = "Rome"
I want the program to output ome.
How can I do that?
This is what I tried:
def getString(x):
return x
def solve(s1, s2):
a = getString(s1[0])
b = getString(s2[0])
for i in range(1, len(s1)):
if s1[i] != s1[i - 1]:
a += getString(s1[i])
for i in range(1, len(s2)):
if s2[i] != s2[i - 1]:
b += getString(s2[i])
if a == b:
print(a)
return True
return False
Edit: gave a wrong answer. This answer works but not the most efficent, although simple
for i in range(len(a)):
if b.endswith(a[i:]):
print(a[i:])
return
I am studying the Bioinformatics course at Coursera, and have been stuck on the following problem for 5 days:
Implement GreedyMotifSearch.
Input: Integers k and t, followed by a collection of strings Dna.
Output: A collection of strings BestMotifs resulting from applying GreedyMotifSearch(Dna, k, t).
If at any step you find more than one Profile-most probable k-mer in a given string, use the
one occurring first.
Here's my attempt to solve this (I just copied it from my IDE, so pardon any print statements):
def GreedyMotifSearch(DNA, k, t):
"""
Documentation here
"""
import math
bestMotifs = []
bestScore = math.inf
for string in DNA:
bestMotifs.append(string[:k])
base = DNA[0]
for i in window(base, k):
newMotifs = []
for j in range(t):
profile = ProfileMatrix([i])
probable = ProfileMostProbable(DNA[j], k, profile)
newMotifs.append(probable)
if Score(newMotifs) <= bestScore:
bestScore = Score(newMotifs)
bestMotifs = newMotifs
return bestMotifs
The helper functions are these:
def SymbolToNumber(Symbol):
"""
Converts base to number (in lexicograpical order)
Symbol: the letter to be converted (str)
Returns: the number correspondinig to that base (int)
"""
if Symbol == "A":
return 0
elif Symbol == "C":
return 1
elif Symbol == "G":
return 2
elif Symbol == "T":
return 3
def NumberToSymbol(index):
"""
Finds base from number (in lexicographical order)
index: the number to be converted (int)
Returns: the base corresponding to index (str)
"""
if index == 0:
return str("A")
elif index == 1:
return str("C")
elif index == 2:
return str("G")
elif index == 3:
return str("T")
def HammingDistance(p, q):
"""
Finds the number of mismatches between 2 DNA segments of equal lengths
p: first DNA segment (str)
q: second DNA segment (str)
Returns: number of mismatches (int)
"""
return sum(s1 != s2 for s1, s2 in zip(p, q))
def window(s, k):
for i in range(1 + len(s) - k):
yield s[i:i+k]
def ProfileMostProbable(Text, k, Profile):
"""
Finds a k-mer that was most likely to be generated by profile among
all k-mers in Text
Text: given DNA segment (str)
k: length of pattern (int)
Profile: a 4x4 matrix (list)
Returns: profile-most probable k-mer (str)
"""
letter = [[] for key in range(k)]
probable = ""
hamdict = {}
index = 1
for a in range(k):
for j in "ACGT":
letter[a].append(Profile[j][a])
for b in range(len(letter)):
number = max(letter[b])
probable += str(NumberToSymbol(letter[b].index(number)))
for c in window(Text, k):
for x in range(len(c)):
y = SymbolToNumber(c[x])
index *= float(letter[x][y])
hamdict[c] = index
index = 1
for pat, ham in hamdict.items():
if ham == max(hamdict.values()):
final = pat
break
return final
def Count(Motifs):
"""
Documentation here
"""
count = {}
k = len(Motifs[0])
for symbol in "ACGT":
count[symbol] = []
for i in range(k):
count[symbol].append(0)
t = len(Motifs)
for i in range(t):
for j in range(k):
symbol = Motifs[i][j]
count[symbol][j] += 1
return count
def FindConsensus(motifs):
"""
Finds a consensus sequence for given list of motifs
motifs: a list of motif sequences (list)
Returns: consensus sequence of motifs (str)
"""
consensus = ""
for i in range(len(motifs[0])):
countA, countC, countG, countT = 0, 0, 0, 0
for motif in motifs:
if motif[i] == "A":
countA += 1
elif motif[i] == "C":
countC += 1
elif motif[i] == "G":
countG += 1
elif motif[i] == "T":
countT += 1
if countA >= max(countC, countG, countT):
consensus += "A"
elif countC >= max(countA, countG, countT):
consensus += "C"
elif countG >= max(countC, countA, countT):
consensus += "G"
elif countT >= max(countC, countG, countA):
consensus += "T"
return consensus
def ProfileMatrix(motifs):
"""
Finds the profile matrix for given list of motifs
motifs: list of motif sequences (list)
Returns: the profile matrix for motifs (list)
"""
Profile = {}
A, C, G, T = [], [], [], []
for j in range(len(motifs[0])):
countA, countC, countG, countT = 0, 0, 0, 0
for motif in motifs:
if motif[j] == "A":
countA += 1
elif motif[j] == "C":
countC += 1
elif motif[j] == "G":
countG += 1
elif motif[j] == "T":
countT += 1
A.append(countA)
C.append(countC)
G.append(countG)
T.append(countT)
Profile["A"] = A
Profile["C"] = C
Profile["G"] = G
Profile["T"] = T
return Profile
def Score(motifs):
"""
Finds score of motifs relative to the consensus sequence
motifs: a list of given motifs (list)
Returns: score of given motifs (int)
"""
consensus = FindConsensus(motifs)
score = 0.0000
for motif in motifs:
score += HammingDistance(consensus, motif)
#print(score)
return round(score, 4)
It seems fine to me. However, when I run this code for quiz problems, it gives an incorrect answer. Their code grading system shows this error:
Failed test #3. Your indexing may be off by one at the beginning of each string in Dna.
I have tried everything I can think of and run this code on all their sample data and debug data, but I simply can't figure out how to make this code work. Please help me with any possible solutions to this.
You have a few problems. I think this should address them all. I've included comments explaining each change along with your original code and a reference to the relevant Pseudocode in the debug data page you linked to.
def GreedyMotifSearch(DNA, k, t):
"""
Documentation here
"""
import math
bestMotifs = []
bestScore = math.inf
for string in DNA:
bestMotifs.append(string[:k])
base = DNA[0]
for i in window(base, k):
# Change here. Should start with one element in motifs and build up.
# As in the line "motifs ← list with only Dna[0](i,k)"
# newMotifs = []
newMotifs = [i]
# Change here to iterate over len(DNA).
# Should go through "for j from 1 to |Dna| - 1"
# for j in range(t):
for j in range(1, len(DNA)):
# Change here. Should build up motifs and build profile using them.
# profile = ProfileMatrix([i])
profile = ProfileMatrix(newMotifs)
probable = ProfileMostProbable(DNA[j], k, profile)
newMotifs.append(probable)
# Change to < rather < = to ensure getting the most recent hit. As referenced in the instructions:
# If at any step you find more than one Profile-most probable k-mer in a given string, use the one occurring **first**.
if Score(newMotifs) < bestScore:
#if Score(newMotifs) <= bestScore:
bestScore = Score(newMotifs)
bestMotifs = newMotifs
return bestMotifs
Have tried searching for this, but can't find exactly what I'm looking for.
I want to make a function that will recursively find the factors of a number; for example, the factors of 12 are 1, 2, 3, 4, 6 & 12.
I can write this fairly simply using a for loop with an if statement:
#a function to find the factors of a given number
def print_factors(x):
print ("The factors of %s are:" % number)
for i in range(1, x + 1):
if number % i == 0: #if the number divided by i is zero, then i is a factor of that number
print (i)
number = int(input("Enter a number: "))
print (print_factors(number))
However, when I try to change it to a recursive function, I am getting just a loop of the "The factors of x are:" statement. This is what I currently have:
#uses recursive function to print all the letters of an integer
def print_factors(x): #function to print factors of the number with the argument n
print ("The factors of %s are:" % number)
while print_factors(x) != 0: #to break the recursion loop
for i in range(1,x + 1):
if x % i == 0:
print (i)
number = int(input("Enter a number: "))
print_factors(number)
The error must be coming in either when I am calling the function again, or to do with the while loop (as far as I understand, you need a while loop in a recursive function, in order to break it?)
There are quite many problems with your recursive approach. In fact its not recursive at all.
1) Your function doesn't return anything but your while loop has a comparision while print_factors(x) != 0:
2) Even if your function was returning a value, it would never get to the point of evaluating it and comparing due to the way you have coded.
You are constantly calling your function with the same parameter over and over which is why you are getting a loop of print statements.
In a recursive approach, you define a problem in terms of a simpler version of itself.
And you need a base case to break out of recursive function, not a while loop.
Here is a very naive recursive approach.
def factors(x,i):
if i==0:
return
if x%i == 0:
print(i)
return factors (x,i-1) #simpler version of the problem
factors(12,12)
I think we do using below method:
def findfactor(n):
factorizeDict
def factorize(acc, x):
if(n%x == 0 and n/x >= x):
if(n/x > x):
acc += [x, n//x]
return factorize(acc, x+1)
else:
acc += [x]
return acc
elif(n%x != 0):
return factorize(acc, x+1)
else:
return acc
return factorize(list(), 1)
def factors(x,i=None) :
if i is None :
print('the factors of %s are : ' %x)
print(x,end=' ')
i = int(x/2)
if i == 0 :
return
if x % i == 0 :
print(i,end=' ')
return factors(x,i-1)
num1 = int(input('enter number : '))
print(factors(num1))
Recursion is a functional heritage and so using it with functional style yields the best results. This means avoiding things like mutations, variable reassignments, and other side effects. That said, here's how I'd write factors -
def factors(n, m = 2):
if m >= n:
return
if n % m == 0:
yield m
yield from factors(n, m + 1)
print(list(factors(10))) # [2,5]
print(list(factors(24))) # [2,3,4,6,8,12]
print(list(factors(99))) # [3,9,11,33]
And here's prime_factors -
def prime_factors(n, m = 2):
if m > n:
return
elif n % m == 0:
yield m
yield from prime_factors(n // m, m)
else:
yield from prime_factors(n, m + 1)
print(list(prime_factors(10))) # [2,5]
print(list(prime_factors(24))) # [2,2,2,3]
print(list(prime_factors(99))) # [3,3,11]
def fact (n , a = 2):
if n <= a :
return n
elif n % a != 0 :
return fact(n , a + 1 )
elif n % a == 0:
return str(a) + f" * {str(fact(n / a , a ))}"
Here is another way. The 'x' is the number you want to find the factors of. The 'c = 1' is used as a counter, using it we'll divide your number by 1, then by 2, all the way up to and including your nubmer, and if the modular returns a 0, then we know that number is a factor, so we print it out.
def factors (x,c=1):
if c == x: return x
else:
if x%c == 0: print(c)
return factors(x,c+1)