NumPy
![]() | |
원저자 | Travis Oliphant |
---|---|
개발자 | Community project |
발표일 | 1995년 (1995)(Numeric), 2006년 (2006)(NumPy) |
안정화 버전 | |
저장소 |
|
프로그래밍 언어 | 파이썬, C |
운영 체제 | 크로스 플랫폼 |
종류 | 수치 해석 |
라이선스 | 수정 BSD 라이선스 |
웹사이트 | www |
NumPy("넘파이"라 읽는다)는 행렬이나 일반적으로 대규모 다차원 배열을 쉽게 처리할 수 있도록 지원하는 파이썬의 라이브러리이다. NumPy는 데이터 구조 외에도 수치 계산을 위해 효율적으로 구현된 기능을 제공한다.
예제
- 배열 생성
>>> import numpy as np >>> x = np.array([1, 2, 3]) >>> x array([1, 2, 3]) >>> y = np.arange(10) # like Python's range, but returns an array >>> y array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
- 기본 작업
>>> a = np.array([1, 2, 3, 6]) >>> b = np.linspace(0, 2, 4) # create an array with four equally spaced points starting with 0 and ending with 2. >>> c = a - b >>> c array([ 1. , 1.33333333, 1.66666667, 4. ]) >>> a**2 array([ 1, 4, 9, 36])
- 유니버셜 함수
>>> a = np.linspace(-np.pi, np.pi, 100) >>> b = np.sin(a) >>> c = np.cos(a)
>>> from numpy.random import rand >>> from numpy.linalg import solve, inv >>> a = np.array([[1, 2, 3], [3, 4, 6.7], [5, 9.0, 5]]) >>> a.transpose() array([[ 1. , 3. , 5. ], [ 2. , 4. , 9. ], [ 3. , 6.7, 5. ]]) >>> inv(a) array([[-2.27683616, 0.96045198, 0.07909605], [ 1.04519774, -0.56497175, 0.1299435 ], [ 0.39548023, 0.05649718, -0.11299435]]) >>> b = np.array([3, 2, 1]) >>> solve(a, b) # solve the equation ax = b array([-4.83050847, 2.13559322, 1.18644068]) >>> c = rand(3, 3) * 20 # create a 3x3 random matrix of values within [0,1] scaled by 20 >>> c array([[ 3.98732789, 2.47702609, 4.71167924], [ 9.24410671, 5.5240412 , 10.6468792 ], [ 10.38136661, 8.44968437, 15.17639591]]) >>> np.dot(a, c) # matrix multiplication array([[ 53.61964114, 38.8741616 , 71.53462537], [ 118.4935668 , 86.14012835, 158.40440712], [ 155.04043289, 104.3499231 , 195.26228855]]) >>> a @ c # Starting with Python 3.5 and NumPy 1.10 array([[ 53.61964114, 38.8741616 , 71.53462537], [ 118.4935668 , 86.14012835, 158.40440712], [ 155.04043289, 104.3499231 , 195.26228855]])
- OpenCV와의 통합
>>> import numpy as np >>> import cv2 >>> r = np.reshape(np.arange(256*256)%256,(256,256)) # 256x256 pixel array with a horizontal gradient from 0 to 255 for the red color channel >>> g = np.zeros_like(r) # array of same size and type as r but filled with 0s for the green color channel >>> b = r.T # transposed r will give a vertical gradient for the blue color channel >>> cv2.imwrite('gradients.png', np.dstack([b,g,r])) # OpenCV images are interpreted as BGR, the depth-stacked array will be written to an 8bit RGB PNG-file called 'gradients.png' True
>>> # # # Pure iterative Python # # # >>> points = [[9,2,8],[4,7,2],[3,4,4],[5,6,9],[5,0,7],[8,2,7],[0,3,2],[7,3,0],[6,1,1],[2,9,6]] >>> qPoint = [4,5,3] >>> minIdx = -1 >>> minDist = -1 >>> for idx, point in enumerate(points): # iterate over all points dist = sum([(dp-dq)**2 for dp,dq in zip(point,qPoint)])**0.5 # compute the euclidean distance for each point to q if dist < minDist or minDist < 0: # if necessary, update minimum distance and index of the corresponding point minDist = dist minIdx = idx >>> print 'Nearest point to q: ', points[minIdx] Nearest point to q: [3, 4, 4] >>> # # # Equivalent NumPy vectorization # # # >>> import numpy as np >>> points = np.array([[9,2,8],[4,7,2],[3,4,4],[5,6,9],[5,0,7],[8,2,7],[0,3,2],[7,3,0],[6,1,1],[2,9,6]]) >>> qPoint = np.array([4,5,3]) >>> minIdx = np.argmin(np.linalg.norm(points-qPoint,axis=1)) # compute all euclidean distances at once and return the index of the smallest one >>> print 'Nearest point to q: ', points[minIdx] Nearest point to q: [3 4 4]
같이 보기
![](http://upload.wikimedia.org/wikipedia/commons/thumb/4/4a/Commons-logo.svg/30px-Commons-logo.svg.png)
- SciPy
- SymPy
- Matplotlib
각주
- ↑ “Numpy News”. 2022년 9월 7일에 확인함.
외부 링크
- NumPy
- 공식 웹사이트
- History of NumPy