Today my task is to make a histogram to represent the operation of A^n where A is a matrix, but only for specific entries in the matrix.
For example, say I have a matrix where the rows sum to one. The first entry is some specific decimal number. However, if I raise that matrix to the 2nd power, that first entry becomes something else, and if I raise that matrix to the 3rd power, it changes again, etc - ad nauseum, and that's what I need to plot.
Right now my attempt is to create an empty list, and then use a for loop to add the entries that result from matrix multiplication to the list. However, all that it does is print the result from the final matrix multiplication into the list, rather than printing its value at each iteration.
Here's the specific bit of code that I'm talking about:
print("The intial probability matrix.")
print(tabulate(matrix))
baseprob = []
for i in range(1000):
matrix_n = numpy.linalg.matrix_power(matrix, s)
baseprob.append(matrix_n.item(0))
print(baseprob)
print("The final probability matrix.")
print(tabulate(matrix_n))
Here is the full code, as well as the output I got.
http://pastebin.com/EkfQX2Hu
Of course it only prints the final value, you are doing the same operation, matrix^s, 1000 times. You need to have s change each of those 1000 times.
If you want to calculate all values in location matrix(0) for matrix^i where i is each value from 1 to s (your final power) do:
baseprob = []
for i in range(1,s): #changed to do a range 1-s instead of 1000
#must use the loop variable here, not s (s is always the same)
matrix_n = numpy.linalg.matrix_power(matrix, i)
baseprob.append(matrix_n.item(0))
Then baseprob will hold matrix(0) for matrix^1, matrix^2, etc. all the way to matrix^s.
Related
I will have an interview with a company which like the hackerearth.com. I don't know how to work and doing the code perfectly. Could you help me with the following example?
This is the example for the .hackerearth.com, however, I don't know that I should consider the constraint in the code? can I use a package like NumPy? or I should only use the basic calculation with my self? Could you check my response and let me know the problem with that? Thank you so much
Input Format:
First line of input consists of an integer N denoting the number of elements in the array A.
Second line consists of N space separated integers denoting the array elements.
Output Format:
The only line of output consists of the value of x.
Input Constraints:
1<N<100
1<A[i]<100
explanation:
An initial sum of array is 1+2+3+4+5=15
When we update all elements to 4, the sum of array which is greater than 15 .
Note that if we had updated the array elements to 3, which is not greater than 15 . So, 4 is the minimum value to which array elements need to be updated.
# Write your code here
import numpy as np
A= [1, 2, 3,4,5]
for i in range(1, max(A)+1):
old = sum(A)
new = sum(i*np.ones(len(A)))
diff = new-old
if diff>0:
print(i)
break
Well this isn't Code Review stack exchange, but:
You don't say how to calculate x. It seems to be something to do with finding an average value, but no-one can judge your code without know what it's trying to do. A web search suggests it is this:
Fredo is assigned a new task today. He is given an array A containing N integers. His task is to update all elements of array to some minimum value x , that is, ; such that sum of this new array is strictly greater than the sum of the initial array. Note that x should be as minimum as possible such that sum of the new array is greater than the sum of the initial array.
Given that the task starts by accepting input, it's important that your program does this part.
N = int(input()) # you can put a prompt string in here, but may conflict with limited output
A = list(map(int,input().split()))
# might need input checks
# might need range checks
# might check that A has exactly N values
you don't need to recalculate old = sum(A) every time around your search loop
calculation of new doesn't need a sum at all - it's just new = i * len(A)
there's no point in checking values of i at or below min(A)
your search will fail if all values of A are the same (try it), because you never look above max(A)
These remarks apply to your approach; a more efficient search would be binary chop, and there is also a mathematical way to go straight to the answer from sum(A) without any searching:
x = sum(A) // len(A) + 1
You don't need numpy or looping for this. Get the average of the array elements, then get the next higher integer from this.
N = 5
A = [1, 2, 3, 4, 5]
total = sum(A)
avg = A/N # not checking for zero-divide because conditions say N > 1
x = floor(avg + 1)
print(x)
Adding 1 is necessary to make the new sum greater than the original sum when the average is an exact integer (e.g. 15/5 == 3).
I have a data variable(sst) in an xarray(nino6), first I use enumerate to assign each value of data variable of the array an index, then I want to calculate with the values of data variable using the index. This code calculates with the indizes itself instead of the data variable values, but I just wanted you to show what I tried.
How can I loop through an index but actually calculating with the values?
for i, entry in enumerate(nino6['sst']):
a=((i-1)+i+(i+1))/3
ssta.append(a)
I apologise for my question is very likely to be really simple (I just started programming), but I searched unsuccesfully here and and on youtube.
If you are trying to get the average of every 3 adjacent numbers in sst, you do it like this:
lst = nino6['sst']
ssta = []
for i in range(1,len(lst) - 1):
a = (lst[i-1] + lst[i] + lst[i+1])/3
ssta.append(a)
Notice that in this implementation, the length of ssta will be smaller than the length of sst by 2 because the first and last numbers do not have flanking numbers. You can have other variations, where you just get the average of two numbers for the first and last numbers.
For alpha and k fixed integers with i < k also fixed, I am trying to encode a sum of the form
where all the x and y variables are known beforehand. (this is essentially the alpha coordinate of a big iterated matrix-vector multiplication)
For a normal sum varying over one index I usually create a 1d array A and set A[i] equal to the i indexed entry of the sum then use sum(A), but in the above instance the entries of the innermost sum depend on the indices in the previous sum, which in turn depend on the indices in the sum before that, all the way back out to the first sum which prevents me using this tact in a straightforward manner.
I tried making a 2D array B of appropriate length and width and setting the 0 row to be the entries in the innermost sum, then the 1 row as the entries in the next sum times sum(np.transpose(B),0) and so on, but the value of the first sum (of row 0) needs to vary with each entry in row 1 since that sum still has indices dependent on our position in row 1, so on and so forth all the way up to sum k-i.
A sum which allows for a 'variable' filled in by each position of the array it's summing through would thusly do the trick, but I can't find anything along these lines in numpy and my attempts to hack one together have thus far failed -- my intuition says there is a solution that involves summing along the axes of a k-i dimensional array, but I haven't been able to make this precise yet. Any assistance is greatly appreciated.
One simple attempt to hard-code something like this would be:
for j0 in range(0,n0):
for j1 in range(0,n1):
....
Edit: (a vectorized version)
You could do something like this: (I didn't test it)
temp = np.ones(n[k-i])
for j in range(0,k-i):
temp = x[:n[k-i-1-j],:n[k-i-j]].T#(y[:n[k-i-j]]*temp)
result = x[alpha,:n[0]]#(y[:n[0]]*temp)
The basic idea is that you try to press it into a matrix-vector form. (note that this is python3 syntax)
Edit: You should note that you need to change the "k-1" to where the innermost sum is (I just did it for all sums up to index k-i)
This is 95% identical to #sehigle's answer, but includes a generic N vector:
def nested_sum(XX, Y, N, alpha):
intermediate = np.ones(N[-1], dtype=XX.dtype)
for n1, n2 in zip(N[-2::-1], N[:0:-1]):
intermediate = np.sum(XX[:n1, :n2] * Y[:n2] * intermediate, axis=1)
return np.sum(XX[alpha, :N[0]] * Y[:N[0]] * intermediate)
Similarly, I have no knowledge of the expression, so I'm not sure how to build appropriate tests. But it runs :\
I have an array D of variable length,
I want to create a loop that performs a sum based on the value of D corresponding to the number of times looped
i.e. the 5th run through the loop would use the 5th value in my array.
My code is:
period = 63 # can be edited to an input() command for variable periods.
Mrgn_dec = .10 # decimal value of 10%, can be manipulated to produce a 10% increase/decrease
rtn_annual = np.arange(0.00,0.15,0.05) # creates an array ??? not sure if helpful
sig_annual = np.arange(0.01,0.31,0.01) #use .31 as python doesnt include the upper range value.
#functions for variables of daily return and risk.
rtn_daily = (1/252)*rtn_annual
sig_daily = (1/(np.sqrt(252)))*sig_annual
D=np.random.normal(size=period) # unsure of range to use for standard distribution
for i in range(period):
r=(rtn_daily+sig_daily*D)
I'm trying to make it so my for loop is multiplied by the value for D of each step.
So D has a random value for every value of period, where period represents a day.
So for the 8th day I want the loop value for r to be multiplied by the 8th value in my array, is there a way to select the specific value or not?
Does the numpy.cumprod command offer any help, I'm not sure how it works but it has been suggested to help the problem.
You can select element in an iterative object (such as D in your code) simply by choosing its index. Such as:
for i in range(period):
print D[i]
But in your code, rtn_daily and sig_daily are not in the same shape, I assume that you want to add sig_daily multiply by D[i] in each position of rtn. so try this:
# -*- coding:utf-8 -*-
import numpy as np
period = 63 # can be edited to an input() command for variable periods.
Mrgn_dec = .10 # decimal value of 10%, can be manipulated to produce a 10% increase/decrease
rtn_annual = np.repeat(np.arange(0.00,0.15,0.05), 31) # creates an array ??? not sure if helpful
sig_annual = np.repeat(np.arange(0.01,0.31,0.01), 3) #use .31 as python doesnt include the upper range value.
#functions for variables of daily return and risk.
rtn_daily = (float(1)/252)*rtn_annual
sig_daily = (1/(np.sqrt(252)))*sig_annual
D=np.random.normal(size=period) # unsure of range to use for standard distribution
print D
for i in range(period):
r=(rtn_daily[i]+sig_daily[i]*D[i])
print r
Last of all, if you are using python2, the division method is for integer, so that means 1/252 will give you zero as result.
a = 1/252 >-- 0
to solve this you may try to make it float:
rtn_daily = (float(1)/252)*rtn_annual
Right now, D is just a scalar.
I'd suggest reading https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.random.normal.html to learn about the parameters.
If you change it to:
D=np.random.normal(mean,stdev,period)
you will get a 1D array with period number of samples, where mean and stdev are your mean and standard deviation of the distribution. Then you change the loop to:
for i in range(period):
r=(rtn_daily+sig_daily*D[i])
EDIT: I don't know what I was thinking when I read the code the first time. It was a horribly bad read on my part.
Looking back at the code, a few things need to happen to make it work.
First:
rtn_annual = np.arange(0.00,0.15,0.05)
sig_annual = np.arange(0.01,0.31,0.01)
These two lines need to be fixed so that the dimensions of the resulting matricies are the same.
Then:
rtn_daily = (1/252)*rtn_annual
Needs to be changed so it doesn't zero everything out -- either change 1 to 1.0 or float(1)
Finally:
r=(rtn_daily+sig_daily*D)
needs to be changed to:
r=(rtn_daily+sig_daily*D[i])
I'm not really sure of the intent of the original code, but it appears as though the loop is unnecessary and you could just change the loop to:
r=(rtn_daily+sig_daily*D[day])
where day is the day you're trying to isolate.
I am fairly new to Python and I am stuck on a particular question and I thought i'd ask you guys.
The following contains my code so far, aswell as the questions that lie therein:
list=[100,20,30,40 etc...]
Just a list with different numeric values representing an objects weight in grams.
object=0
while len(list)>0:
list_caluclation=list.pop(0)
print(object number:",(object),"evaluates to")
What i want to do next is evaluate the items in the list. So that if we go with index[0], we have a list value of 100. THen i want to separate this into smaller pieces like, for a 100 gram object, one would split it into five 20 gram units. If the value being split up was 35, then it would be one 20 gram unit, on 10 gram unit and one 5 gram unit.
The five units i want to split into are: 20, 10, 5, 1 and 0.5.
If anyone has a quick tip regarding my issue, it would be much appreciated.
Regards
You should think about solving this for a single number first. So what you essentially want to do is split up a number into a partition of known components. This is also known as the Change-making problem. You can choose a greedy algorithm for this that always takes the largest component size as long as it’s still possible:
units = [20, 10, 5, 1, 0.5]
def change (number):
counts = {}
for unit in units:
count, number = divmod(number, unit)
counts[unit] = count
return counts
So this will return a dictionary that maps from each unit to the count of that unit required to get to the target number.
You just need to call that function for each item in your original list.
One way you could do it with a double for loop. The outer loop would be the numbers you input and the inner loop would be the values you want to evaluate (ie [20,10,5,1,0.5]). For each iteration of the inner loop, find how many times the value goes into the number (using the floor method), and then use the modulo operator to reassign the number to be the remainder. On each loop you can have it print out the info that you want :) Im not sure exactly what kind of output you're looking for, but I hope this helps!
Ex:
import math
myList=[100,20,30,40,35]
values=[20,10,5,1,0.5]
for i in myList:
print(str(i)+" evaluates to: ")
for num in values:
evaluation=math.floor(i/num)
print("\t"+str(num)+"'s: "+str(evaluation))
i%=num