I tried using for-loop to exec my function but its always the same result.
import random
def main(arr):
result = random.choice(arr)
...some code...
return len(result)
for i in range(100):
main(arr)
I could only get diff result from stop/run the terminal. Anyone know why?
my question is the same as this one. random.choice always same
import random
results = []
with open('kargerMinCut.txt') as inputfile:
for line in inputfile:
results.append(line.strip().split('\t'))
def contract(arr):
while len(arr) > 2:
# Generate random number in list of lists
# ranList = random.choice(arr)
ranList = arr[np.random.choice(len(arr))]
ranNum = random.choice(ranList[1:])
# retrieve the index of the random number
listIndex = arr.index(ranList)
for i in range(0, len(arr)):
if arr[i][0] == ranNum:
targetList = i
break
target = ranList[0]
for i in range(0, len(arr)):
if i == listIndex:
arr[i].pop(0)
arr[i] = [x for x in arr[i] if x != ranNum]
elif i == targetList:
arr[i] = [x for x in arr[i] if x != target]
else:
for index, item in enumerate(arr[i]):
if item == target:
arr[i][index] = ranNum
arr[targetList] += arr[listIndex]
del arr[listIndex]
return len(arr[0])-1
the arr would be like this
array = [[1,2,3,4],[2,1,3,4],[3,1,2,4],[4,1,2,3]]
I don't know what you do inside your function but I've got the normal result. And in the question what you linked to the person just used seed. This is kinda pseudorandom that gives you all the time the same random output. Here is the link to deep your knowledge about pseudorandom
import random
arr = [1,2,3,4,5]
def main(arr):
result = random.choice(arr)
print(result)
for i in range(100):
main(arr)
The result is as it has to be:
1
3
5
3
4
3
1
4
4
3
2
Related
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.
I've drafted the below code for the captioned question, but the return result is always 0. Could anyone please help me figure out what's the problem here?
Thanks a lot!
import random
dice_sides = 6
frequency_list = []
def roll_dice(times):
results = []
for roll_num in range(times):
result = random.randint(1,dice_sides)
results.append(result)
for i in range(dice_sides):
if results.count(i) != 2:
frequency = 0
break
else:
frequency = 1
return frequency
def occurrence(N,times):
for j in range(N):
frequency_list.append(roll_dice(times))
prob = frequency_list.count(1)
return prob
print(occurrence(10000,12))
You can try something like this
Code
import random
from collections import Counter
def roll_dice(n_sides, times):
if n_sides % times:
return 0
results = []
for roll_num in range(times):
result = random.randint(1, n_sides)
results.append(result)
# I'm using set here and will check its length,
# Counter(results) returns a dict of items (item, count)
# and if every item has the same count it should have length 1.
# More generic statement not only for (2 in this case)
res_dict = set(Counter(results).values())
if len(res_dict) == 1:
return 1
return 0
def mean(ar):
return sum(ar)/len(ar)
def occurrence(N, n_sides, times):
frequency_list = []
for j in range(N):
frequency_list.append(roll_dice(n_sides, times))
prob = mean(frequency_list)
return prob
if __name__ == '__main__':
N = 100000 # I intentionally made it 100k
n_sides = 6
times = 12
res_prob = occurrence(N, times)
print(res_prob)
Output
0.00604
[Finished in 3.6s]
I am new to coding and so the following code I wrote may be incorrect or sub-optimal. However, the problem I have is that I do not understand the input and thus cannot get the code to run (I only tested it with custom inputs).
The essence of the problem is that you have some sequence of numbers and you want to arrange the sequence monotonically (nondecreasing or nonincreasing). You do this by a random shuffle. How many shuffles does it take for you to get to the monotonic sequence via a random shuffle? You can find the problem here and here is my code below:
#!/bin/python3 ------ The following import is given in the prompt
import os
import sys
# Complete the solve function below. Here is my code below
def solve(P):
P.sort()
correct = P
count = []
i = 0
# Here I am trying to calculate the number of ways to get the desired monotonic sequence
# I count the number of repeated numbers in the sequence as separate items in a list
for j in range(i,len(correct)):
if correct[i] != correct[j] or i == len(correct) - 1:
count.append(j-i)
i = j
j = len(correct)
else:
j = j + 1
summ = 0
for k in range(len(count)):
summ = summ + count[k]
if summ == len(correct):
i = len(correct)
poss = [1]*len(count)
for k in range(len(count)):
for l in range(1,count[k]+1):
poss[k] = poss[k]*l
possible = 1
for x in poss:
possible = possible * x
# This is calculating the number of different permutations of n distinct elements
total = 1
n = len(correct)
for i in range(1,n+1):
total = total * i
# Calculating the probability to the 6th decimal place
probability = float(possible / total)
expected = round(1/probability, 6)
print(expected)
# The following code is given in the prompt to input the test cases
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
P_count = int(input())
P = list(map(int, input().rstrip().split()))
result = solve(P)
fptr.write(str(result) + '\n')
fptr.close()
In my code, I just assumed that P is the sequence of numbers you receive from the input.
I have to solve how to replace the elements below zero elements with zeros and output the sum of the remaining elements in the matrix.
For example, [[0,3,5],[3,4,0],[1,2,3]] should output the sum of 3 + 5 + 4 + 1 + 2, which is 15.
So far:
def matrixElementsSum(matrix):
out = 0
# locate the zeros' positions in array & replace element below
for i,j in enumerate(matrix):
for k,l in enumerate(j):
if l == 0:
break
out += l
return out
The code outputs seemingly random numbers.
Can someone fix what's wrong? Thanks
You can easily drop elements that are below a zero element is by using the zip function.
def matrixElementsSum(matrix):
out = 0
# locate the zeros' positions in array & replace element below
for i,j in enumerate(matrix):
# elements in the first row cannot be below a '0'
if i == 0:
out += sum(j)
else:
k = matrix[i-1]
for x, y in zip(j, k):
if y != 0:
out += x
return out
Now consider naming your variables a little more meaningfully. Something like:
def matrixElementsSum(matrix):
out = 0
# locate the zeros' positions in array & replace element below
for row_number, row in enumerate(matrix):
# elements in the first row cannot be below a '0'
if row_number == 0:
out += sum(row)
else:
row_above = matrix[row_number - 1]
for element, element_above in zip(row, row_above):
if element_above != 0:
out += element
return out
You should look into list comprehensions to make the code even more readable.
Python questions again.
I want to count the number of comparison operations performed by quick sort. Because I use a recursive function, I do not think that assigning count = 0 to the beginning of the function body is inappropriate, so I made it as follows.
def QuickSort(lst, count = 0):
if len(lst) > 1:
pivot_idx = len(lst) // 2
smaller_nums, larger_nums = [], []
for idx, num in enumerate(lst):
if idx != pivot_idx:
if num < lst[pivot_idx]:
smaller_nums.append(num)
else:
larger_nums.append(num)
count = QuickSort(smaller_nums, count + 1)[1]
count = QuickSort(larger_nums, count + 1)[1]
lst[:] = smaller_nums + [lst[pivot_idx]] + larger_nums
return lst, count
However, after counting, I confirmed the count which is much lower than my expectation. According to big o, the quick sort would have to show the calculation of n * log (n), but it showed a much lower count. For example, when sorting a list with 1000 random elements, we expected to see a count of 1000 * log (1000) = 6907, but actually only 1164 counts. I am wondering if I am misusing the count in the function or misunderstanding it.
Thank you.
Your post is mistaken on several points:
Big-O is allows arbitrary constant factors and also ignoring the values for "small" values of n, where "small" can be arbitrarily large for any given analysis. So your computations are meaningless.
Your counts are wrong. There's one comparison per loop iteration. You're counting something else.
This is a strange way to code the count. Just use a global variable.
Try this. Note really you're using twice as many comparisons as this reports. The check that the loop index isn't the pivot could be eliminated with a smarter implementation.
c = 0
def QuickSort(lst):
if len(lst) <= 1:
return lst
pivot_idx = len(lst) // 2
smaller, larger = [], []
for idx, num in enumerate(lst):
if idx != pivot_idx:
global c
c += 1
(larger, smaller)[num < lst[pivot_idx]].append(num)
return QuickSort(smaller) + [lst[pivot_idx]] + QuickSort(larger)
def Run(n):
lst = [random.randint(0,1000) for r in xrange(n)]
QuickSort(lst)
print c
Run(1000)
If you're aghast at the prospect of using a global variable, then you can just wrap the sort in a class:
import random
class QuickSort:
def __init__(self):
self.comparisons = 0
def sort(self, lst):
if len(lst) <= 1:
return lst
pivot_idx = len(lst) // 2
smaller, larger = [], []
for idx, num in enumerate(lst):
if idx != pivot_idx:
self.comparisons += 1
(larger, smaller)[num < lst[pivot_idx]].append(num)
return self.sort(smaller) + [lst[pivot_idx]] + self.sort(larger)
def Run(n):
lst = [random.randint(0,1000) for r in xrange(n)]
quicksort = QuickSort()
print quicksort.sort(lst)
print quicksort.comparisons
Run(100)
Building on the answer provided by Gene by adding print statements and a sort "error" range, his example was very helpful to my understanding of quicksort and an error term on the big O impact of operations performance comparison.
class QuickSort:
def __init__(self):
self.comparisons = 0
def sort(self, lst):
k_err = 0 # k << n, the value the sort array can be in error
if len(lst) <= 1:
return lst
pivot_idx = len(lst) // 2
smaller, larger = [], []
for idx, num in enumerate(lst):
if idx != (pivot_idx) :
self.comparisons += 1
try:
(larger, smaller)[(num - k_err) < lst[pivot_idx]].append(num)
except:
(larger, smaller)[(num + k_err) < lst[pivot_idx]].append(num)
print(pivot_idx,"larger", self.comparisons, larger)
print(pivot_idx, "smaller", self.comparisons, smaller, )
return self.sort(smaller) + [lst[pivot_idx]] + self.sort(larger)
def Run(n):
random.seed(100)
lst = [random.randint(0,round(100,0)) for r in range(n)]
quicksort = QuickSort()
print(len(lst), lst)
print(quicksort.sort(lst))
print(quicksort.comparisons, quicksort.comparisons/n, ((quicksort.comparisons/n)/math.log(n,10)), math.log(n,10) )