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Need help. How I can get sine square in python? Is there specific method or function?
Some supposedly obvious solutions are NOT suitable. Examples:
import numpy as np
import math
x = np.arange(0, 3, 0.5)
print([(math.sin(i) ** 2) for i in x])
print([math.sin(math.sin(i))for i in x])
# [0.0, 0.22984884706593015, 0.7080734182735712, 0.9949962483002227, 0.826821810431806, 0.3581689072683868]
#[0.0, 0.4612695550331807, 0.7456241416655579, 0.8401148815567654, 0.7890723435728884, 0.5633808209655248]
# or
x = np.arange(0, 3, 0.5)
print(np.sin(x) ** 2)
print(np.sin(np.sin(x)))
# [0. 0.22984885 0.70807342 0.99499625 0.82682181 0.35816891]
# [0. 0.46126956 0.74562414 0.84011488 0.78907234 0.56338082]
You need to look for math module in Python. See this.
math.sin(x) ** 2
You can also use math.pow(x,y). See this for how to raise a number x raised to the power y.
A small example program.
import math
rad = int(input("Enter radians: "))
print(math.sin(rad) ** 2)
If you want to convert from radians to degrees or vice-versa, have a look at this.
You will be safer if you use it as follows -
((math.sin(x)) ** 2)
Emphasis is to put the math.sin(x) inside bracket.
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I was trying to create this [80000, 104000, 135000...] list in Python. Its the value, starting at 80,000 multiplied by 1.3 each time I want
What i've tried:
a = [num*1.5 for num in ??? if num>=80000] #???--> i've tried range(10)
I should be able to do this but I can't find any solutions rn..
I must use list-comprehensions, if possible.
Some help would be nice, thank you!
There is a very basic mathematical operation that represents multiplying by the same value many time: power.
a = [80000 * (1.3**n) for n in range(100)]
You could write your own generator then use that in conjunction with a list comprehension.
def numgen(start, factor, limit):
for _ in range(limit):
yield int(start)
start *= factor
mylist = [value for value in numgen(80_000, 1.3, 10)]
print(mylist)
Output:
[80000, 104000, 135200, 175760, 228488, 297034, 386144, 501988, 652584, 848359]
import numpy as np
print(80000 * 1.3**np.arange(3))
# [ 80000. 104000. 135200.]
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I'd like to make a function that rounds integers or floats homogeneously and smart. For example, if I have an array like:
[0.672, 0.678, 0.672]
my output would be:
[0.67, 0.68, 0.67]
but also if I have this kind of input:
[17836.982, 160293.673, 103974.287]
my output would be:
[17836, 160293, 103974]
But at the same time, if my array only has close together values such as:
[17836.987, 17836.976, 17836.953]
The output would be:
[17836.99, 17836.98, 17836.95]
An automated way could be to compute all absolute differences, getting the min and finding out the number of decimal places to keep to maintain a representative difference.
This doesn't give the exact output you want but follows the general logic.
Here using numpy to help on the computation, the algorithm is O(n**2):
def auto_round(l, round_int_part=False):
import numpy as np
a = np.array(l)
b = abs(a-a[:,None])
np.fill_diagonal(b, float('inf'))
n = int(np.ceil(-np.log10(b.min())))
# print(f'rounding to {n} decimals') # uncomment to get info
if n<0:
if not round_int_part:
return a.astype(int).tolist()
return np.round(a, decimals=n).astype(int).tolist()
return np.round(a, decimals=n).tolist()
auto_round([17836.987, 17836.976, 17836.953])
# [17836.99, 17836.98, 17836.95]
auto_round([0.6726, 0.6785, 0.6723])
# [0.6726, 0.6785, 0.6723]
auto_round([17836.982, 160293.673, 103974.287])
# [ 17836, 160293, 103974]
auto_round([17836.982, 160293.673, 103974.287], round_int_part=True)
# [20000, 160000, 100000]
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I am trying to implement this summation, any help, please?
enter image description here
Considering that x is a list of numbers, this can be written as following:
H = '+'.join([f'x{i}*x{i+1}' for i in range(1, 21)])
>>>print(H)
'x1*x2+x2*x3+x3*x4+x4*x5+x5*x6+x6*x7+x7*x8+x8*x9+x9*x10+x10*x11+x11*x12+x12*x13+x13*x14+x14*x15+x15*x16+x16*x17+x17*x18+x18*x19+x19*x20+x20*x21'
You need a computer algebra system (CAS). There are a number of software packages in Python that implements CAS, for example, using sympy:
from sympy import *
i = symbols('i')
x = IndexedBase('x')
H = Sum(x[i] * x[i-1], (i, 1, 20))
This will produce H, a symbolic expression the represents the equation in your image.
Or, you can use summation instead to evaluate the Sum into a series of additions:
H_evaluated = summation(x[i] * x[i-1], (i, 1, 20))
or call the doit() function:
H_evaluated == H.doit()
You can try sympy online on https://live.sympy.org/ to play with sympy without installing it locally. Try this equation live
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I have this code in MATLAB and I am trying to convert it in Python.
A=[1,-0.75,0.25]
yc(1:45)=-2;
y(1:6)=0
u(1:6)=0
[lig,col]=size(A);
alpha(1)=1;
alpha(2)=A(2)-1;
if(col>2)
for i=3:col
alpha(i)=A(i)-A(i-1);
end ;
end;
alpha(col+1)=-A(col);
I don't know how to convert it in python thnx for helping me
It would be better if your code would have been a minimal example of what you are trying to do. You are defining variables that are not even used. But here's a more or less literal translation. Note that you probably want to preallocate alpha (both in Matlab and Python)
import numpy as np
A = np.array([1.0, -.75, .25])
yc = -2 * np.ones(45)
y = np.zeros(6)
u = np.zeros(6)
col = A.size
alpha = np.array([1, A[1] - 1])
if col > 2:
for i in range(2, col):
alpha = np.append(alpha, A[i] - A[i-1])
alpha = np.append(alpha, -A[col-1])
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I am trying to identify what the problem with the differentiation of trig functions in Python. I use scipy.misc.derivative.
Correct case:
def f(x):
return math.sin(x)
y=derivative(f,5.0,dx=1e-9)
print(y)
This will give a math.cos(5) right?
My problem is here. Since python accepts radians, we need to correct what is inside the sin function. I use math.radians.
If I code it again:
def f(x):
return math.sin(math.radians(x))
y=derivative(f,5.0,dx=1e-9)
print(y)
This will give an answer not equal to what I intended which should be math.cos(math.radians(5)).
Am i missing something?
You have to be consistent with the argument of the trigonometric function. Is not that "Python accepts radians", all programming languages I know use radians by default (including Python).
If you want to get the derivative of 5 degrees, yes, first convert to radians and then use it as the argument of the trigonometric function. Obviously, when you do
y=derivative(f,5.0,dx=1e-9)
using
def f(x):
return math.sin(x)
you get f'(x)=cos(x) evaluated at 5 (radians). If you want to check that the result is correct this is the function to check, not f'(x)=cos(math.radians(x)), which will give you another result.
If you want to pass 5 degrees, yes, you will need to get the radians first:
y=derivative(f,math.radians(5.0),dx=1e-9)
which will be the same as cos(math.radians(5)).
Here is a working example
from scipy.misc import derivative
import math
def f(x):
return math.sin(x)
def dfdx(x):
return math.cos(x)
y1 = derivative(f,5.0,dx=1e-9)
y2 = dfdx(5)
print(y1) # 0.28366
print(y2) # 0.28366