based on this function. I'm trying to create two empty arrays (one for x and other for y), which later I will use to plot in python. But before anything this is what I have so far...
import math
x1=-2.0
x2=2.0
arr1 = []
arr2 = []
i=0
n=10
delta=(x2-x1)/n
for i in range (0,n+1):
x=x1+delta*i
arr1.append(x)
print arr1
# I have not called the w function yet
the code above creates a list of 10 numbers for now to keep it simple. Then it will send the elements of the array to the function below and compute the equation with certain numbers(infinite loop).
#This function will create the array list for y
import math
def w(x, limit):# the limit here to compare when the number is really small
suma = 0.0
sumb = 0.0
m=1
x=0
suma=suma+((1/(math.pow(2,m))*(math.sin(math.pow(2,m)*x)))
sumb=suma+((1/(math.pow(2,m+1))*(math.sin(math.pow(2,m+1)*x))) # I'm having a
#syntax error
#here
x+=0
if (abs (suma-sumb)<limit):
break:
else m+=1:
if (m<20):
break:
I will appreciate any help with my syntax errors or any suggestion. I just hope I was clear enough.
Thanks ahead of time
The syntax error is actually on the previous line, where the parenthesis are not balanced. You need an extra ) at the end of that line (and at the one you indicated as giving an error too btw).
There are also a few other issues
suma is set to zero, so suma = suma + ... is the same as suma = ..., but I'm guessing you still need to add while loop before this line.
On the line indicated, you have sumb = suma +, which is probably a copy/paste mistake.
The code block starting at x+=0 is indented by only 3 spaces instead of 4. This is probably not the case in your actual code, but if it is, Python will complain about that too.
else m+=1: should be else: m+=1 (colon directly after else, not at the end of the line.
break: should just be break (without to colon).
Related
Is there a simple and quick way to multiply a column of a matrix with element of a vector. We can do this explicitly,
program test
integer :: x(3,3), y(3), z(3,3)
x = reshape([(i,i=1,9)],[3,3])
y = [1,2,3]
do i=1,3
z(:,i) = x(:,i) * y(i)
print *, z(:,i)
enddo
end program test
Is there a way to perform the do loop in one line. For example in Numpy python we can do this to do the job in one shot
z = np.einsum('ij,i->ij',x,y)
#or
z = x*y[:,None]
Try
z = x * spread(y,1,3)
and if that doesn't work (no Fortran on this computer so I haven't checked) fiddle around with spread until it does. In practice you'll probably want to replace the 3 by size(x,1) or suchlike.
I expect that this will cause the compiler to create temporary arrays. And I expect it will be easy to find situations where it underperforms the explicit looping scheme in the question. 'neat' one-liners often have a cost in both time and space. And often tried-and-trusted Fortran approach of explicit looping is the one to go with.
Why replace clear easy to read code with garbage?
program test
implicit none
integer i,j
integer :: x(3,3), y(3), z(3,3)
x = reshape([(i,i=1,9)],[3,3])
y = [1,2,3]
z = reshape ([((x(j,i)*y(i) ,j=1,3),i=1,3)], [3,3])
print *, z(1,:)
print *, z(2,:)
print *, z(3,:)
end program test
I got a numpy.ndarray of electromagnetic samples as complex numbers, where the format is as follows:
ex1:
[[ 8.23133235e-15, -1.59200901e-15, -4.39818917e-13, 7.68089585e-13]
[ 6.98151957e-15, -1.20306059e-15, 9.83923013e-13, 1.64838108e-11]
[ 8.41053742e-15, -1.77702007e-15, -5.98961364e-13, 8.97436205e-13]
[ 7.08443026e-15, -1.25262430e-15, 1.11415868e-12, 1.69346186e-11]]
where rows make up real and imaginary part alternately:
[[z1Ex.real, z1Ey.real, z1Hx.real, z1Hy.real],
[z1Ex.imag, z1Ey.imag, z1Hx.imag, z1Hy.imag],
[z2Ex.real, z2Ey.real, z2Hx.real, z2Hy.real],
[z2Ex.imag, z2Ey.imag, z2Hx.imag, z2Hy.imag],
...etc.]
What I want is to create a new array which expresses the data in magnitude and phase, but keep the same format (i.e. replace real rows with magnitude rows and imaginary with phase rows).
I managed to put up list comprehensions for both calculations (which I´m fairly proud of, being an 2-week amateur, so please be gentle;)). The result for magnitude is what I´d expect, but the phase is terribly off and I don´t have any idea why...
My approach:
Slice the original array in real and imag sub-arrays:
import numpy, cmath
real = ex1[::2] #numpy.ndarray
imag = ex1[1::2] #numpy.ndarray
Define lambdas outside of list comprehension:
magcalc = lambda z, y: abs(complex(z, y))
phasecalc = lambda z,y: cmath.phase(complex(z, y))
Define list comprehension to do math on sub-arrays:
real[:] = np.array([[magcalc(z,y) for z, y in zip(real[x],imag[x])] for x in xrange(len(real))])
imag[:] = np.array([[phasecalc(z,y) for z, y in zip(real[x],imag[x])] for x in xrange(len(imag))])
Check results in original array:
print ex1[:4]
If I do that, the phase result for the first Ex sample is 0.574 rad. If I check the phase manually (i.e. cmath.phase(complex(z1Ex.real,z1Ex.imag))), then I get 0.703 rad. I would accept if there was smth wrong in my list comprehensions, but the magnitude results are completely correct, so I doubt that that´s it.
Where am I doing it wrong? I really tried to find out for 2 days straight now, no luck... Also, I can´t think of another way to achieve what I want.
Please help... (Using Python 2.7)
Thanks
Nils
Oh jeez.. Now I saw the problem, can´t believe how dense I am... Credit goes to John, for making me re-think variable assignments.
In imag[:] = np.array([[phasecalc(z,y) for z, y in zip(real[x],imag[x])] for x in xrange(len(imag))]), I refer to real[], as if it was still populated with real values. But I changed real[] the line before to contain magnitude... So, just changing the variable names for the list comprehensions will do it:
Define list comprehension to do math on sub-arrays:
realcopy[:] = np.array([[magcalc(z,y) for z, y in zip(real[x],imag[x])] for x in xrange(len(real))])
imagcopy[:] = np.array([[phasecalc(z,y) for z, y in zip(real[x],imag[x])] for x in xrange(len(imag))])
And then re-assign to original mag, phase arrays:
Check original results
real[:] = realcopy
imag[:] = imagcopy
print ex1[:4]
Sorry for the waste of time and bytes...
Cheers
Nils
I am having a problem with an xor search.
I have an array composed of binary values. My list contains 1000 distinct binary values, and I want to time how long it takes for a double loop to find an element in the list. Therefore for a double loop search, I expect it to go through the loop [(1) + (2) +(3)+...+(1000)] = 500500 times. [n(n+1) / 2]
I use the bitwise_xor in the following code
from numpy import bitwise_xor
count = 0
for word1 in listOutTextnoB:
for word2 in listOutTextnoB:
count+=1
if bitwise_xor(word1,word2)==0:
break
print "count"
Unfortunately, when I print count, I get count = 1,000,000
If I change the if statement to
if bitwise_xor(word1,word2):
break
count is 1000
I also tried to do:
if word1^word2==0:
break
but it gives me "TypeError: unsupported operand type(s) for ^: 'str' and 'str'"
A working example would be:
1101110111010111011101101110110010111100101111001 XOR 1101110111010111011101101110110010111100101111001
it should give me 0 and exit the inner loop
What is wrong with code?
^ works on integers, not arrays, so that is not surprising.
I don't know why you used strings but:
from numpy import bitwise_xor
listOutTextnoB = range(1000)
count = 0
for word1 in listOutTextnoB:
for word2 in listOutTextnoB:
count+=1
if bitwise_xor(word1,word2)==0:
break
print "count", count
prints
count 500500
as you predict.
EDIT: yes, you should be doing
if int(word1) ^ int(word2) == 0:
break
bitwise_xor is actually returning 'NotImplemented' for every string, string input.
Your error shows the problem: the values in your list are strings, not numbers. I'm not sure what bitwise_xor does to them, but I'm pretty sure it won't convert them to numbers first. If you do this manually (bitwise_xor (int (word1), int (word2))), I think it should work.
I want to limit the size of a list in python 2.7 I have been trying to do it with a while loop but it doesn't work
l=[]
i=raw_input()//this is the size of the list
count=0
while count<i:
l.append(raw_input())
count=count+1
The thing is that it does not finish the loop. I think this problem has an easy answer but I can't find it.
Thanks in advance
I think the problem is here:
i=raw_input()//this is the size of the list
raw_input() returns a string, not an integer, so comparisons between i and count don't make sense. [In Python 3, you'd get the error message TypeError: unorderable types: int() < str(), which would have made things clear.] If you convert i to an int, though:
i = int(raw_input())
it should do what you expect. (We'll ignore error handling etc. and possibly converting what you're adding to l if you need to.)
Note though that it would be more Pythonic to write something like
for term_i in range(num_terms):
s = raw_input()
l.append(s)
Most of the time you shouldn't need to manually keep track of indices by "+1", so if you find yourself doing it there's probably a better way.
That is because i has a string value type, and int < "string" always returns true.
What you want is:
l=[]
i=raw_input() #this is the size of the list
count=0
while count<int(i): #Cast to int
l.append(raw_input())
count=count+1
You should try changing your code to this:
l = []
i = input() //this is the size of the list
count = 0
while count < i:
l.append(raw_input())
count+=1
raw_input() returns a string while input() returns an integer. Also count+=1 is better programming practice than count = count + 1. Good luck
In my c++ class, we got assigned pairs. Normally I can come up with an effective algorithm quite easily, this time I cannot figure out how to do this to save my life.
What I am looking for is someone to explain an algorithm (or just give me tips on what would work) in order to get this done. I'm still at the planning stage and want to get this code done on my own in order to learn. I just need a little help to get there.
We have to create histograms based on a 4 or 5 integer input. It is supposed to look something like this:
Calling histo(5, 4, 6, 2) should produce output that appears like:
*
* *
* * *
* * *
* * * *
* * * *
-------
A B C D
The formatting to this is just killing me. What makes it worse is that we cannot use any type of arrays or "advanced" sorting systems using other libraries.
At first I thought I could arrange the values from highest to lowest order. But then I realized I did not know how to do this without using the sort function and I was not sure how to go on from there.
Kudos for anyone who could help me get started on this assignment. :)
Try something along the lines of this:
Determine the largest number in the histogram
Using a loop like this to construct the histogram:
for(int i = largest; i >= 1; i--)
Inside the body of the loop, do steps 3 to 5 inclusive
If i <= value_of_column_a then print a *, otherwise print a space
Repeat step 3 for each column (or write a loop...)
Print a newline character
Print the horizontal line using -
Print the column labels
Maybe i'm mistaken on your q, but if you know how many items are in each column, it should be pretty easy to print them like your example:
Step 1: Find the Max of the numbers, store in variable, assign to column.
Step 2: Print spaces until you get to column with the max. Print star. Print remaining stars / spaces. Add a \n character.
Step 3: Find next max. Print stars in columns where the max is >= the max, otherwise print a space. Add newline. at end.
Step 4: Repeat step 3 (until stop condition below)
when you've printed the # of stars equal to the largest max, you've printed all of them.
Step 5: add the -------- line, and a \n
Step 6: add row headers and a \n
If I understood the problem correctly I think the problem can be solved like this:
a= <array of the numbers entered>
T=<number of numbers entered> = length(a) //This variable is used to
//determine if we have finished
//and it will change its value
Alph={A,B,C,D,E,F,G,..., Z} //A constant array containing the alphabet
//We will use it to print the bottom row
for (i=1 to T) {print Alph[i]+" "}; //Prints the letters (plus space),
//one for each number entered
for (i=1 to T) {print "--"}; //Prints the two dashes per letter above
//the letters, one for each
while (T!=0) do {
for (i=1 to N) do {
if (a[i]>0) {print "*"; a[i]--;} else {print " "; T--;};
};
if (T!=0) {T=N};
}
What this does is, for each non-zero entered number, it will print a * and then decrease the number entered. When one of the numbers becomes zero it stops putting *s for its column. When all numbers have become zero (notice that this will occur when the value of T comes out of the for as zero. This is what the variable T is for) then it stops.
I think the problem wasn't really about histograms. Notice it also doesn't require sorting or even knowing the