clamp

Pythonic way to replace list values with upper and lower bound (clamping, clipping, thresholding)?

落爺英雄遲暮 提交于 2019-12-09 05:04:32
问题 I want to replace outliners from a list. Therefore I define a upper and lower bound. Now every value above upper_bound and under lower_bound is replaced with the bound value. My approach was to do this in two steps using a numpy array. Now I wonder if it's possible to do this in one step, as I guess it could improve performance and readability. Is there a shorter way to do this? import numpy as np lowerBound, upperBound = 3, 7 arr = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) arr[arr >

forcing the columns of a matrix within different limits

╄→尐↘猪︶ㄣ 提交于 2019-12-07 18:44:48
问题 I have a matrix named l having size 20X3. What I wanted to do was this : Suppose I have this limits: l1_max=20; l1_min=0.5; l2_max=20; l2_min=0.5; mu_max=20; mu_min=0.5; I wanted to force all the elements of the matrix l within the limits. The values of 1st column within l1_max & l1_min. The values of 2nd column within l2_max & l2_min. The values of 3rd column within mu_max & mu_min. What I did was like this: for k=1:20 if l(k,1)>l1_max l(k,1) = l1_max; elseif l(k,1)<l1_min l(k,1) = l1_min;

forcing the columns of a matrix within different limits

此生再无相见时 提交于 2019-12-06 10:54:58
I have a matrix named l having size 20X3. What I wanted to do was this : Suppose I have this limits: l1_max=20; l1_min=0.5; l2_max=20; l2_min=0.5; mu_max=20; mu_min=0.5; I wanted to force all the elements of the matrix l within the limits. The values of 1st column within l1_max & l1_min. The values of 2nd column within l2_max & l2_min. The values of 3rd column within mu_max & mu_min. What I did was like this: for k=1:20 if l(k,1)>l1_max l(k,1) = l1_max; elseif l(k,1)<l1_min l(k,1) = l1_min; end if l(k,2)>l2_max l(k,2) = l2_max; elseif l(k,2)<l2_min l(k,2) = l2_min; end if l(k,3)>mu_max l(k,3)

Pythonic way to replace list values with upper and lower bound (clamping, clipping, thresholding)?

丶灬走出姿态 提交于 2019-12-03 04:44:41
I want to replace outliners from a list. Therefore I define a upper and lower bound. Now every value above upper_bound and under lower_bound is replaced with the bound value. My approach was to do this in two steps using a numpy array. Now I wonder if it's possible to do this in one step, as I guess it could improve performance and readability. Is there a shorter way to do this? import numpy as np lowerBound, upperBound = 3, 7 arr = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) arr[arr > upperBound] = upperBound arr[arr < lowerBound] = lowerBound # [3 3 3 3 4 5 6 7 7 7] print(arr) You can use numpy

Limit integer to bounds [duplicate]

倾然丶 夕夏残阳落幕 提交于 2019-12-01 09:41:00
This question already has an answer here: Most efficient/elegant way to clip a number? 12 answers I'm trying to make sure that int x is greater or equal than 0 but smaller than 1080 (screen size in this case). I came up with this int x = 123; x = std::min(std::max(x, 0), 1080); This seems ugly. Is there a better way to achieve this? If you live in the future, you can use std::clamp from C++17 : x = std::clamp(x, 0, 1080); Naive solution looks ok too: int x = 123; if (x < 0) x = 0; if (x > 1080) x = 1080; or wrap everything in a function: template<typename T> T between(const T x, const T min,

Limit integer to bounds [duplicate]

北慕城南 提交于 2019-12-01 08:49:29
问题 This question already has answers here : Most efficient/elegant way to clip a number? (12 answers) Closed last year . I'm trying to make sure that int x is greater or equal than 0 but smaller than 1080 (screen size in this case). I came up with this int x = 123; x = std::min(std::max(x, 0), 1080); This seems ugly. Is there a better way to achieve this? 回答1: If you live in the future, you can use std::clamp from C++17: x = std::clamp(x, 0, 1080); 回答2: Naive solution looks ok too: int x = 123;

Problems limiting object rotation with Mathf.Clamp()

谁说胖子不能爱 提交于 2019-11-30 16:04:10
I am working on a game that rotates an object on the z axis. I need to limit the total rotation to 80 degrees. I tried the following code, but it doesn't work. minAngle = -40.0f and maxAngle = 40.0f Vector3 pos = transform.position; pos.z = Mathf.Clamp(pos.z, minAngle, maxAngle); transform.position = pos; The code you posted clamps the z position. What you want is to use transform.rotation void ClampRotation(float minAngle, float maxAngle, float clampAroundAngle = 0) { //clampAroundAngle is the angle you want the clamp to originate from //For example a value of 90, with a min=-45 and max=45,

pytorch clamp 与clamp_区别

[亡魂溺海] 提交于 2019-11-30 10:50:20
pytorch clamp 与clamp_ ,有下划线的表示修改并付给自身,无下划线的表示需要返回处理后的值,比如: h = k.clamp(min=0) #将结果存入h,k保留原值 k.clamp_(min=0) # 将结果存入k 来源: https://www.cnblogs.com/YouXiangLiThon/p/11579895.html

Clamping floating numbers in Python?

ぐ巨炮叔叔 提交于 2019-11-30 00:20:06
问题 Is there a built-in function for this in Python 2.6? Something like: clamp(myValue, min, max) 回答1: There's no such function, but max(min(my_value, max_value), min_value) will do the trick. 回答2: Numpy's clip function will do this. >>> import numpy >>> numpy.clip(10,0,3) 3 >>> numpy.clip(-4,0,3) 0 >>> numpy.clip(2,0,3) 2 回答3: I think the question is answered but here's an alternative DIY solution if anyone needs it: def clip(value, lower, upper): return lower if value < lower else upper if