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I have this
for A in [0, -0.25, 0.25, -0.5, 0.5, -0.75, 0.75, -1.0, 1.0, -1.25, 1.25, -1.5, 1.5, -1.75, 1.75, -2.0, 2.0, -2.25, 2.25, -2.5, 2.5, -2.75, 2.75, -3.0, 3.0, -3.25, 3.25, -3.5, 3.5, -3.75, 3.75, -4.0, 4.0, -4.25, 4.25, -4.5, 4.5, -4.75, 4.75, -5.0, 5.0]:
Is it possible to make it with generator function? I have now this:
def frange(start, stop, step=1.0):
while start <= stop:
yield start
start += step
and use like this:
for error in self.frange(-2.5, 2.5, 0.25):
but its returns [-2.5, 2.25, ... , 0 , 2.25, 2.5] and for my program it's very hard to calculate. because I finding the value the near to zero, but I don't know how much combinations it could be.
I need go from zero and next value must be in minus and plus value.
like [0, -0.25, 0.25...].
Maybe you meant a generator instead of a lambda:
def opposing_numbers(increment, maximum):
yield 0
value = increment
while value <= maximum:
yield -value
yield value
value += increment
Then call it as:
opposing_numbers(0.25, 5)
You could use the numpy.arange() function, and then sort the values by their absolute value:
answer = sorted(np.arange(-5, 5.25, 0.25), key=abs)
print(answer)
Output
[0.0,
-0.25,
0.25,
-0.5,
0.5,
...,
-4.5,
4.5,
-4.75,
4.75,
-5.0,
5.0]
Try this:
[(i // 2) * 0.25 * (2*(i&1)- 1) for i in range(1,42)]
here is a cool one liner
from itertools import chain
list(chain(*zip([i /4.0 for i in range(10)], [-i/4.0 for i in range(9)])))
[0.0, 0.0, 0.25, -0.25, 0.5, -0.5, 0.75, -0.75, 1.0, -1.0, 1.25, -1.25, 1.5, -1.5, 1.75, -1.75, 2.0, -2.0]
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I have an array with positive and negative numbers and want to do a cumulative sum of numbers of the same sign until the next number carries an opposite sign. It starts again at 0. Maybe better explained with a sample.
Here is the original array:
np.array([0.2, 0.5, 1.3, 0.6, -0.3, -1.1, 0.2, -2.0, 0.7, 1.1, 0.0, -1.2])
And the output I expect without using a loop, of course:
np.array([0.0, 0.0, 0.0, 2.6, 0.0, -1.4, 0.2, -2.0, 0.0, 0.0, 1.8, -1.2])
Any efficient idea would help a lot...
One vectorial option:
a = np.array([0.2, 0.5, 1.3, 0.6, -0.3, -1.1, 0.2, -2.0, 0.7, 1.1, 0.0, -1.2])
cs = np.cumsum(a)
idx = np.nonzero(np.r_[np.diff(a>0), True])
out = np.zeros_like(a)
out[idx] = np.diff(np.r_[0, cs[idx]])
Output:
array([ 0. , 0. , 0. , 2.6, 0. , -1.4, 0.2, -2. , 0. , 1.8, 0. , -1.2])
As seen here How do I convert a Python list into a C array by using ctypes? this code will take a python array and transform it to a C array.
import ctypes
arr = (ctypes.c_int * len(pyarr))(*pyarr)
Which would the way of doing the same with a list of lists or a lists of lists of lists?
For example, for the following variable
list3d = [[[40.0, 1.2, 6.0, 0.3], [50.0, 4.2, 0, 0]], [[40.0, 1.2, 6.0, 0.3], [50.0, 4.2, 0, 0]], [[40.0, 1.2, 6.0, 0.3], [50.0, 4.2, 0, 0]]]
I have tried the following with no luck:
([[ctypes.c_double * 4] *2]*3)(*list3d)
# *** TypeError: 'list' object is not callable
(ctypes.c_double * 4 *2 *3)(*list3d)
# *** TypeError: expected c_double_Array_4_Array_2 instance, got list
Thank you!
EDIT: Just to clarify, I am trying to get one object that contains the whole multidimensional array, not a list of objects. This object's reference will be an input to a C DLL that expects a 3D array.
It works with tuples if you don't mind doing a bit of conversion first:
from ctypes import *
list3d = [
[[0.0, 1.0, 2.0, 3.0], [4.0, 5.0, 6.0, 7.0]],
[[0.2, 1.2, 2.2, 3.2], [4.2, 5.2, 6.2, 7.2]],
[[0.4, 1.4, 2.4, 3.4], [4.4, 5.4, 6.4, 7.4]],
]
arr = (c_double * 4 * 2 * 3)(*(tuple(tuple(j) for j in i) for i in list3d))
Check that it's initialized correctly in row-major order:
>>> (c_double * 24).from_buffer(arr)[:]
[0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0,
0.2, 1.2, 2.2, 3.2, 4.2, 5.2, 6.2, 7.2,
0.4, 1.4, 2.4, 3.4, 4.4, 5.4, 6.4, 7.4]
Or you can create an empty array and initialize it using a loop. enumerate over the rows and columns of the list and assign the data to a slice:
arr = (c_double * 4 * 2 * 3)()
for i, row in enumerate(list3d):
for j, col in enumerate(row):
arr[i][j][:] = col
I made the change accordingly
a = [[[40.0, 1.2, 6.0, 0.3], [50.0, 4.2, 0, 0]], [[40.0, 1.2, 6.0, 0.3], [50.0, 4.2, 0, 0]], [[40.0, 1.2, 6.0, 0.3], [50.0, 4.2, 0, 0]]]
arr = (((ctypes.c_float * len(a[0][0])) * len(a[0])) * len(a))
arr_instance=arr()
for i in range(0,len(a)):
for j in range(0,len(a[0])):
for k in range(0,len(a[0][0])):
arr_instance[i][j][k]=a[i][j][k]
The arr_instance is what you want.
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I want to calculate percentile values for 10%, 50% and 90%. So the inputs would be a percentile you want to find and an array of values to calculate. How would I do this? It's been a while since stats...
Help in powershell or python would be appreciated.
Edit: Sorry, I meant creating my own function rather than using a pre built function/library
You can do it using numpy in the following way:
import numpy as np
a = np.array([1,2,3,4,5])
p = np.percentile(a, 50)
You can read more about the percentile function in the attached link.
Other option is to use statistics.quantiles this will give you a distribution list of n - 1 cut points separating the n quantile intervals.
Examlpe of use:
from statistics import quantiles
quantiles([1, 2, 3, 4, 5], n=100)
# [0.06, 0.12, 0.18, 0.24, 0.3, 0.36, 0.42, 0.48, 0.54, 0.6, 0.66, 0.72, 0.78, 0.84, 0.9, 0.96, 1.02, 1.08, 1.14, 1.2, 1.26, 1.32, 1.38, 1.44, 1.5, 1.56, 1.62, 1.68, 1.74, 1.8, 1.86, 1.92, 1.98, 2.04, 2.1, 2.16, 2.22, 2.28, 2.34, 2.4, 2.46, 2.52, 2.58, 2.64, 2.7, 2.76, 2.82, 2.88, 2.94, 3.0, 3.06, 3.12, 3.18, 3.24, 3.3, 3.36, 3.42, 3.48, 3.54, 3.6, 3.66, 3.72, 3.78, 3.84, 3.9, 3.96, 4.02, 4.08, 4.14, 4.2, 4.26, 4.32, 4.38, 4.44, 4.5, 4.56, 4.62, 4.68, 4.74, 4.8, 4.86, 4.92, 4.98, 5.04, 5.1, 5.16, 5.22, 5.28, 5.34, 5.4, 5.46, 5.52, 5.58, 5.64, 5.7, 5.76, 5.82, 5.88, 5.94]
quantiles([1, 2, 3, 4, 5], n=100)[49]
Edit
To create your own function please refer to the following link: https://code.activestate.com/recipes/511478-finding-the-percentile-of-the-values/
You can find percentile with numpy
import numpy as np
arr = [20, 2, 7, 1, 34]
percentile_arr = [10,50,90]
for i in range(0,len(percentile_arr)):
percentile = np.percentile(arr, percentile_arr[i])
print(f"{percentile_arr[i]}th percentile of array is : {percentile}")
Edit
You can find different approaches with and without numpy here
Hmmm, by an array you meant a list?
If that's so, then you have very good option: for loop
my_values = [...]
result = []
percentage = 0,5
for i in my_values:
result.append(i*percentage)
the append method of the list result is an way of telling python "hey, I want you to add this thing over here on the list"
I am trying to check if a numpy array contains a specific value:
>>> x = np.linspace(-5,5,101)
>>> x
array([-5. , -4.9, -4.8, -4.7, -4.6, -4.5, -4.4, -4.3, -4.2, -4.1, -4. ,
-3.9, -3.8, -3.7, -3.6, -3.5, -3.4, -3.3, -3.2, -3.1, -3. , -2.9,
-2.8, -2.7, -2.6, -2.5, -2.4, -2.3, -2.2, -2.1, -2. , -1.9, -1.8,
-1.7, -1.6, -1.5, -1.4, -1.3, -1.2, -1.1, -1. , -0.9, -0.8, -0.7,
-0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0. , 0.1, 0.2, 0.3, 0.4,
0.5, 0.6, 0.7, 0.8, 0.9, 1. , 1.1, 1.2, 1.3, 1.4, 1.5,
1.6, 1.7, 1.8, 1.9, 2. , 2.1, 2.2, 2.3, 2.4, 2.5, 2.6,
2.7, 2.8, 2.9, 3. , 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7,
3.8, 3.9, 4. , 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8,
4.9, 5. ])
>>> -5. in x
True
>>> a = 0.2
>>> a
0.2
>>> a in x
False
I assigned a constant to variable a. It seems that the precision of a is not compatible with the elements in the numpy array generated by np.linspace().
I've searched the docs, but didn't find anything about this.
This is not a question of the precision of np.linspace, but rather of the type of the elements in the generated array.
np.linspace generates elements which, conceptually, equally divide the input range between them. However, these elements are then stored as floating point numbers with limited precision, which makes the generation process itself appear to lack precision.
By passing the dtype argument to np.linspace, you can specify the precision of the floating point type used to store its result, which can increase the apparent precision of the generation process.
Nevertheless, you should not use the equality operator to compare floating point numbers. Instead, use np.isclose in conjunction with np.ndarray.any, or some equivalent:
>>> floats_64 = np.linspace(-5, 5, 101, dtype='float64')
>>> floats_128 = np.linspace(-5, 5, 101, dtype='float128')
>>> print(0.2 in floats_64)
False
>>> print(floats_64[52])
0.20000000000000018
>>> print(np.isclose(0.2, floats_64).any()) # check if any element in floats_64 is close to 0.2
True
>>> print(0.2 in floats_128)
False
>>> print(floats_128[52])
0.20000000000000017764
>>> print(np.isclose(0.2, floats_128).any()) # check if any element in floats_128 is close to 0.2
True
I have a coding interface which has a counter component. It simply increments by 1 with every update. Consider it an infinite generator of {1,2,3,...} over time which I HAVE TO use.
I need to use this value and iterate from -1.5 to 1.5. So, the iteration should start from -1.5 and reach 1.5 and then from 1.5 back to -1.5.
How should I use this infinite iterator to generate an iteration in that range?
You can use cycle from itertools to repeat a sequence.
from itertools import cycle
# build the list with 0.1 increment
v = [(x-15)/10 for x in range(31)]
v = v + list(reversed(v))
cv = cycle(v)
for c in my_counter:
x = next(cv)
This will repeat the list v:
-1.5, -1.4, -1.3, -1.2, -1.1, -1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4,
-0.3, -0.2, -0.1, 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0,
1.1, 1.2, 1.3, 1.4, 1.5, 1.5, 1.4, 1.3, 1.2, 1.1, 1.0, 0.9, 0.8, 0.7,
0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0.0, -0.1, -0.2, -0.3, -0.4, -0.5, -0.6,
-0.7, -0.8, -0.9, -1.0, -1.1, -1.2, -1.3, -1.4, -1.5, -1.5, -1.4, -1.3,
-1.2, -1.1, -1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1,
0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3,
1.4, 1.5, 1.5, 1.4, 1.3, 1.2, 1.1, 1.0, 0.9 ...
Something like:
import itertools
infGenGiven = itertools.count() # This is similar your generator
def func(x):
if x%2==0:
return 1.5
else:
return -1.5
infGenCycle = itertools.imap(func, infGenGiven)
count=0
while count<10:
print infGenCycle.next()
count+=1
Output:
1.5
-1.5
1.5
-1.5
1.5
-1.5
1.5
-1.5
1.5
-1.5
Note that this starts 1.5 because the first value in infGenGiven is 0, although for your generator it is 1 and so the infGenCycle output will give you what you want.
Thank you all.
I guess the best approach is to use the trigonometric functions (sine or cosine) which oscillate between plus and minus one.
More details at: https://en.wikipedia.org/wiki/Trigonometric_functions
Cheers