I want to generate a list of lists that contains a progressive number of randomly generated binary values.
How do I add a condition that tells python to add random values to a list until it reaches a specified length? In this case, the length of each new list should be a progressively larger odd number.
from random import randint
shape = []
odds = [x for x in range(100) if x % 2 == 1]
while len(shape) < 300:
for x in odds:
randy = [randint(0,1)] * x ?? # need to run this code x amount of times
shape.append(randy) # so that each len(randy) = x
*I would prefer to not use count += 1
desired output:
shape
[[0],[0,1,0],[1,1,0,1,0],[1,0,0,0,1,1,0]...etc]
You want a generator expression list comprehension:
randy = [randint(0, 1) for i in range(x)]
The problem with [someFunc()] * someNum is that Python first evaluates the inner expression, someFunc() and resolves it to some number before executing the outer expression.
Related
#create a simple list in python
#list comprehension
x = [i for i in range(100)]
print (x)
#using loops
squares = []
for x in range(10):
squares.append(x**2)
print (squares)
multiples = k*[z for z in x] for k in squares
So in the last line of code I am trying to multiply both the lists. the problem is the lists are not of the same side and k*[z for z in x] this part is also incorrect.
For problems with iteration, I suggest anyone to check Loop Like A Native by Ned Batchelder and Looping like a Pro by David Baumgold
Option 1
If you want to multiply them as far as the shortest list goes, zip is your friend:
multiples = [a * b for a, b in zip (x, squares)]
Option 2
If you want a matrix with the product, then you can do it like this
result = [
[a * b for a in x]
for b in squares
]
I don't quite understand what the desired output would be. As the function stands now, you would have a list of lists, where the first element has 100 elements, the second one 400, the third 900, and so on.
One thing that's strange: The expression [z for z in x] defines a list that is identical to x. So, you might just write k*x
If you want to multiply the elements of both lists, you would have to write [[k*z for z in x] for k in squares]. This would lead to a list of 10 lists of 100 elements (or a 10x100-matrix) containing the products of your lists.
If you want to have one list of length 100 in the end that holds some kind of products, you will have to think about how to proceed with the shorter list.
EDIT: Or if you want to multiply them as far as possible until you reach the end of the shorter list, FRANCOIS CYRIL's solution is an elegant way to do so.
You can loop on each array to multiply element by element at the same position in a result array:
i = 0
arrayRes = []
while i < min (len(array1),len(array2)):
arrayRes.append(array1[i]*array2[i])
i+=1
Or do you prefer to multiply them, matrix way?
x = 0
y = 0
arrayRes = []
while x < len(array1):
arrayRes.append([])
while y < len(array2):
arrayRes[x].append(array1[x]*array2[y])
y+=1
x+=1
i do learn Python for scientific working. At the moment i try to generate a 10x10 random Matrix with binary entries: 0 and 1. I already got a solution with numpy BUT im interested of what is the error in my own solution.
The Idea is to Access every entry of my Matrix seperately and assign a value to it by calling random.randint(0, 1 ) within two while loops. In Advance i define a dummy 10x10 Matrix called "World" and reassign ist values it in the loop. The Code Looks how follows:
import random
World=list(10*[10*[0]]) #this is my dummy matrix
i=0
j=0
while i <= 9:
while j <= 9:
World[i][j]=random.randint(0, 1) #here i want to Access a specific element of my dummy Matrix and "overwrite" it
if j == 9:
j=0 #if the counter "j" reaches 9 - the last element - it shall assign j=0 and leave the innermost while loop by "break"
break
j=j+1
i=i+1
for x in World:
print(*x)
The Problem with the Output should be obvious:
columns are equal
I am hopefully u understand what was my Intention here and can help me fix my code. I tried many many Things but i did not fix this.
I already found a 2-line short solution which i will use in my final Code but i want to run this also on my own because i am convinced this could work also well.
Many Thanks in Advance.
- Wendel
Your error is in the creation of the list.
NOTE:
[0] * m returns just a reference to a list of m zeros, but not a list.
The subsequent repeating of this element creates a list of n items
that all reference to the same list (just as well as the operation b =
a for lists does not create the new list), so all rows in the
resulting list are actually the same string.
import random
#World=list(10*[10*[0]]) #this is my dummy matrix
n = 10
World= [0] * n
for i in range(n):
World[i] = [0] * n
i=0
j=0
while i <= 9:
while j <= 9:
World[i][j]=random.randint(0, 1) #here i want to Access a specific element of my dummy Matrix and "overwrite" it
if j == 9:
j=0 #if the counter "j" reaches 9 - the last element - it shall assign j=0 and leave the innermost while loop by "break"
break
j=j+1
i=i+1
for x in World:
print(*x)
Suppose that two numbers are given: the number of rows of n and the number of columns m. You must create a list of size n×m, filled with, say, zeros.
The obvious solution appears to be wrong:
a = [[0] * m] * n
This can be easily seen if you set the value of a[0][0] to 5, and then print the value of a[1][0] — it will also be equal to 5. The reason is, [0] * m returns just a reference to a list of m zeros, but not a list. The subsequent repeating of this element creates a list of n items that all reference to the same list (just as well as the operation b = a for lists does not create the new list), so all rows in the resulting list are actually the same string.
n = 3
m = 4
a = [[0] * m] * n
a[0][0] = 5
print(a[1][0])
A possible way: you can create a list of n elements (say, of n zeros) and then make each of the elements a link to another one-dimensional list of m elements:
n = 3
m = 4
a = [0] * n
for i in range(n):
a[i] = [0] * m
Another (but similar) way: create an empty list and then append a new element to it n times (this element should be a list of length m):
n = 3
m = 4
a = []
for i in range(n):
a.append([0] * m)
But the easiest way is to use generator, creating a list of n elements, each of which is a list of m zeros:
n = 3
m = 4
a = [[0] * m for i in range(n)]
In this case each element is created independently from the others. The list [0] * m is n times consructed as the new one, and no copying of references occurs.
This question already has answers here:
How to most efficiently increase values at a specified range in a large array and then find the largest value
(5 answers)
Closed 5 years ago.
You are given a list of size N, initialized with zeroes. You have to perform M operations on the list and output the maximum of final values of all the N elements in the list. For every operation, you are given three integers a,b and k and you have to add value to all the elements ranging from index to (both inclusive).
Input Format
First line will contain two integers N and M separated by a single space.
Next M lines will contain three integers a,b and k separated by a single space.
Numbers in list are numbered from 1 to N .
Constraints
Click here
Output Format
A single line containing maximum value in the updated list.
Sample Input
5 3
1 2 100
2 5 100
3 4 100
Sample Output
200
Explanation
After first update list will be 100 100 0 0 0.
After second update list will be 100 200 100 100 100.
After third update list will be 100 200 200 200 100.
So the required answer will be 200.
One of the Solutions with less time complexity
n, inputs = [int(n) for n in input().split(" ")]
list = [0]*(n+1)
for _ in range(inputs):
x, y, incr = [int(n) for n in input().split(" ")]
list[x-1] += incr
if((y)<=len(list)):
list[y] -= incr
max = x = 0
for i in list:
x=x+i;
if(max<x):max=x
print(max)
Can someone explain the above solution?
Basically it stores deltas rather than the final list; that means each operation only takes 2 reads and writes rather than (b - a + 1). Then the final max scan adds the deltas as it goes along, which is still an O(n) operation which you would have had to do anyway.
n, inputs = [int(n) for n in input().split(" ")]
Get the list size (n) and number of operations (m), ie 5 and 3
list = [0]*(n+1)
Create an empty 0-filled list. Should be lst = [0] * n (do not use list as a variable name, it shadows the built-in type) (we do not need an extra end cell, except as a checksum on our algorithm - if it works properly the final checksum should be 0).
for _ in range(inputs):
x, y, incr = [int(n) for n in input().split(" ")]
Get an operation (a, b, k) ie 1, 2, 100.
list[x-1] += incr
Add delta to the starting cell
if((y)<=len(list)):
list[y] -= incr
Subtract the delta from the ending cell (should be if y < n: lst[y] -= incr)
The algorithm may be easier to understand if you add a print(lst) here (after the if but inside the for loop).
Now process the deltas to find the maximum item:
max = x = 0
for i in list:
x=x+i;
x is now the value the actual value of the current list cell. Also max is a terrible variable name because it shadows the built-in max() function.
if(max<x):max=x
Keep a running max tally
print(max)
Show the result.
I'm trying to implement Radix sort in python.
My current program is not working correctly in that a list like [41,51,2,3,123] will be sorted correctly to [2,3,41,51,123], but something like [52,41,51,42,23] will become [23,41,42,52,51] (52 and 51 are in the wrong place).
I think I know why this is happening, because when I compare the digits in the tens place, I don't compare units as well (same for higher powers of 10).
How do I fix this issue so that my program runs in the fastest way possible? Thanks!
def radixsort(aList):
BASEMOD = 10
terminateLoop = False
temp = 0
power = 0
newList = []
while not terminateLoop:
terminateLoop = True
tempnums = [[] for x in range(BASEMOD)]
for x in aList:
temp = int(x / (BASEMOD ** power))
tempnums[temp % BASEMOD].append(x)
if terminateLoop:
terminateLoop = False
for y in tempnums:
for x in range(len(y)):
if int(y[x] / (BASEMOD ** (power+1))) == 0:
newList.append(y[x])
aList.remove(y[x])
power += 1
return newList
print(radixsort([1,4,1,5,5,6,12,52,1,5,51,2,21,415,12,51,2,51,2]))
Currently, your sort does nothing to reorder values based on anything but their highest digit. You get 41 and 42 right only by chance (since they are in the correct relative order in the initial list).
You should be always build a new list based on each cycle of the sort.
def radix_sort(nums, base=10):
result_list = []
power = 0
while nums:
bins = [[] for _ in range(base)]
for x in nums:
bins[x // base**power % base].append(x)
nums = []
for bin in bins:
for x in bin:
if x < base**(power+1):
result_list.append(x)
else:
nums.append(x)
power += 1
return result_list
Note that radix sort is not necessarily faster than a comparison-based sort. It only has a lower complexity if the number of items to be sorted is larger than the range of the item's values. Its complexity is O(len(nums) * log(max(nums))) rather than O(len(nums) * log(len(nums))).
Radix sort sorts the elements by first grouping the individual digits of the same place value. [2,3,41,51,123] first we group them based on first digits.
[[],[41,51],[2],[3,123],[],[],[],[],[],[]]
Then, sort the elements according to their increasing/decreasing order. new array will be
[41,51,2,3,123]
then we will be sorting based on tenth digit. in this case [2,3]=[02,03]
[[2,3],[],[123],[],[41],[51],[],[],[],[]]
now new array will be
[2,3,123,41,51]
lastly based on 100th digits. this time [2,3,41,51]=[002,003,041,051]
[[2,3,41,51],[123],[],[],[],[],[],[],[],[]]
finally we end up having [2,3,41,51,123]
def radixsort(A):
if not isinstance(A,list):
raise TypeError('')
n=len(A)
maxelement=max(A)
digits=len(str(maxelement)) # how many digits in the maxelement
l=[]
bins=[l]*10 # [[],[],.........[]] 10 bins
for i in range(digits):
for j in range(n): #withing this we traverse unsorted array
e=int((A[j]/pow(10,i))%10)
if len(bins[e])>0:
bins[e].append(A[j]) #adds item to the end
else:
bins[e]=[A[j]]
k=0 # used for the index of resorted arrayA
for x in range(10):#we traverse the bins and sort the array
if len(bins[x])>0:
for y in range(len(bins[x])):
A[k]=bins[x].pop(0) #remove element from the beginning
k=k+1
I am working with python 3.2 and I spent a lot of time trouble shooting this, and I still can't seem to wrap my brain around it.
number = random.randint ( x0 ,xn )
I'm generating a random number. It's purpose is to make my code come at me differently everytime.
For example I have 10 variables of text that I have written. I have solved the problem of not having these variables appear in the same order at each program run.
The issue I have is that they now appear randomly everytime. It picks one out of 10 everytime, instead the first time 10 and next 9. I can't seem to find out how to exclude the previous ones.
thelist = [0]
while i < x
if number in thelist:
>>>repeat<<<
else:
thelist.append (number)
if ( number == x0 ):
>>>something<<<
elif ( number == x1 ):
>>>something<<<
This is what I would imagine the code would look like, everytime you loop one more number gets appended to the list, so that everytime it picks a number already in the list it repeats the loop again until it then has used all the numbers that random.randint can pull.
Here's a shuffle function:
import random
max = 15
x = list(range(max+1))
for i in range(max, 0, -1):
n = random.randint(0, i)
x[n], x[i] = x[i], x[n]
This starts with a sorted list of numbers [0, 1, ... max].
Then, it chooses a number from index 0 to index max, and swaps it with index max.
Then, it chooses a number from index 0 to index max-1, and swaps it with index max-1.
And so on, for max-2, max-3, ... 1
As yosukesabai rightly notes, this has the same effect as calling random.sample(range(max+1), max+1). This picks max + 1 unique random values from range(max+1). In other words, it just shuffles the order around. Docs: http://docs.python.org/2/library/random.html#random.sample
If you wanted something more along the lines of your proposed algorithm, you could do:
import random
max = 15
x = range(max+1)
l = []
for _ in range(max+1):
n = random.randint(0,max)
while n in l:
n = random.randint(0,max)
l.append(n)
From what I understand of your description and sample code, you want thelist to end up with every integer between x0 and xn in a random order. If so, you can achieve that very simply with random.shuffle(), which shuffles a list in place:
import random
x0 = 5
xn = 15
full_range = list(range(x0, xn))
print(full_range)
random.shuffle(full_range)
print(full_range)