Find the number which will replace the in the given sequence 1/2, 3, 7 15 24 51 67
Prerequisites#You’ll need to know a bit of Python. For a refresher, see the Python tutorial. Show
To work the examples, you’ll need Learner profile This is a quick overview of arrays in NumPy. It demonstrates how n-dimensional (\(n>=2\)) arrays are represented and can be manipulated. In particular, if you don’t know how to apply common functions to n-dimensional arrays (without using for-loops), or if you want to understand axis and shape properties for n-dimensional arrays, this article might be of help. Learning Objectives After reading, you should be able to:
The Basics#NumPy’s main object is the homogeneous multidimensional array. It is a table of elements (usually numbers), all of the same type, indexed by a tuple of non-negative integers. In NumPy dimensions are called axes. For example, the array for the coordinates of a point in 3D space, [[1., 0., 0.], [0., 1., 2.]] NumPy’s array class is called the number of axes (dimensions) of the array. ndarray.shapethe dimensions of the array. This is a tuple of integers indicating the size of
the array in each dimension. For a matrix with n rows and m columns, the total number of elements of the array. This is equal to the product of the elements of an object describing the type of the elements in the array. One can create or specify dtype’s using standard Python types. Additionally NumPy provides types of its own. numpy.int32, numpy.int16, and numpy.float64 are some examples. ndarray.itemsizethe size in bytes of each element of the array. For example, an array of elements of type the buffer containing the actual elements of the array. Normally, we won’t need to use this attribute because we will access the elements in an array using indexing facilities. An example#>>> import numpy as np >>> a = np.arange(15).reshape(3, 5) >>> a array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14]]) >>> a.shape (3, 5) >>> a.ndim 2 >>> a.dtype.name 'int64' >>> a.itemsize 8 >>> a.size 15 >>> type(a) Array Creation#There are several ways to create arrays. For example, you can create an array from a regular Python list or tuple using the >>> import numpy as np >>> a = np.array([2, 3, 4]) >>> a array([2, 3, 4]) >>> a.dtype dtype('int64') >>> b = np.array([1.2, 3.5, 5.1]) >>> b.dtype dtype('float64') A frequent error consists in calling >>> a = np.array(1, 2, 3, 4) # WRONG Traceback (most recent call last): ... TypeError: array() takes from 1 to 2 positional arguments but 4 were given >>> a = np.array([1, 2, 3, 4]) # RIGHT
>>> b = np.array([(1.5, 2, 3), (4, 5, 6)]) >>> b array([[1.5, 2. , 3. ], [4. , 5. , 6. ]]) The type of the array can also be explicitly specified at creation time: >>> c = np.array([[1, 2], [3, 4]], dtype=complex) >>> c array([[1.+0.j, 2.+0.j], [3.+0.j, 4.+0.j]]) Often, the elements of an array are originally unknown, but its size is known. Hence, NumPy offers several functions to create arrays with initial placeholder content. These minimize the necessity of growing arrays, an expensive operation. The function >>> np.zeros((3, 4)) array([[0., 0., 0., 0.], [0., 0., 0., 0.], [0., 0., 0., 0.]]) >>> np.ones((2, 3, 4), dtype=np.int16) array([[[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]], [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]], dtype=int16) >>> np.empty((2, 3)) array([[3.73603959e-262, 6.02658058e-154, 6.55490914e-260], # may vary [5.30498948e-313, 3.14673309e-307, 1.00000000e+000]]) To create sequences of numbers, NumPy provides the >>> np.arange(10, 30, 5) array([10, 15, 20, 25]) >>> np.arange(0, 2, 0.3) # it accepts float arguments array([0. , 0.3, 0.6, 0.9, 1.2, 1.5, 1.8]) When >>> from numpy import pi >>> np.linspace(0, 2, 9) # 9 numbers from 0 to 2 array([0. , 0.25, 0.5 , 0.75, 1. , 1.25, 1.5 , 1.75, 2. ]) >>> x = np.linspace(0, 2 * pi, 100) # useful to evaluate function at lots of points >>> f = np.sin(x) See also
Printing Arrays#When you print an array, NumPy displays it in a similar way to nested lists, but with the following layout:
One-dimensional arrays are then printed as rows, bidimensionals as matrices and tridimensionals as lists of matrices. >>> a = np.arange(6) # 1d array >>> print(a) [0 1 2 3 4 5] >>> >>> b = np.arange(12).reshape(4, 3) # 2d array >>> print(b) [[ 0 1 2] [ 3 4 5] [ 6 7 8] [ 9 10 11]] >>> >>> c = np.arange(24).reshape(2, 3, 4) # 3d array >>> print(c) [[[ 0 1 2 3] [ 4 5 6 7] [ 8 9 10 11]] [[12 13 14 15] [16 17 18 19] [20 21 22 23]]] See below to get more details on If an array is too large to be printed, NumPy automatically skips the central part of the array and only prints the corners: >>> print(np.arange(10000)) [ 0 1 2 ... 9997 9998 9999] >>> >>> print(np.arange(10000).reshape(100, 100)) [[ 0 1 2 ... 97 98 99] [ 100 101 102 ... 197 198 199] [ 200 201 202 ... 297 298 299] ... [9700 9701 9702 ... 9797 9798 9799] [9800 9801 9802 ... 9897 9898 9899] [9900 9901 9902 ... 9997 9998 9999]] To disable this behaviour
and force NumPy to print the entire array, you can change the printing options using >>> np.set_printoptions(threshold=sys.maxsize) # sys module should be imported Basic Operations#Arithmetic operators on arrays apply elementwise. A new array is created and filled with the result. >>> a = np.array([20, 30, 40, 50]) >>> b = np.arange(4) >>> b array([0, 1, 2, 3]) >>> c = a - b >>> c array([20, 29, 38, 47]) >>> b**2 array([0, 1, 4, 9]) >>> 10 * np.sin(a) array([ 9.12945251, -9.88031624, 7.4511316 , -2.62374854]) >>> a < 35 array([ True, True, False, False]) Unlike
in many matrix languages, the product operator >>> A = np.array([[1, 1], ... [0, 1]]) >>> B = np.array([[2, 0], ... [3, 4]]) >>> A * B # elementwise product array([[2, 0], [0, 4]]) >>> A @ B # matrix product array([[5, 4], [3, 4]]) >>> A.dot(B) # another matrix product array([[5, 4], [3, 4]]) Some operations, such as >>> rg = np.random.default_rng(1) # create instance of default random number generator >>> a = np.ones((2, 3), dtype=int) >>> b = rg.random((2, 3)) >>> a *= 3 >>> a array([[3, 3, 3], [3, 3, 3]]) >>> b += a >>> b array([[3.51182162, 3.9504637 , 3.14415961], [3.94864945, 3.31183145, 3.42332645]]) >>> a += b # b is not automatically converted to integer type Traceback (most recent call last): ... numpy.core._exceptions._UFuncOutputCastingError: Cannot cast ufunc 'add' output from dtype('float64') to dtype('int64') with casting rule 'same_kind' When operating with arrays of different types, the type of the resulting array corresponds to the more general or precise one (a behavior known as upcasting). >>> a = np.ones(3, dtype=np.int32) >>> b = np.linspace(0, pi, 3) >>> b.dtype.name 'float64' >>> c = a + b >>> c array([1. , 2.57079633, 4.14159265]) >>> c.dtype.name 'float64' >>> d = np.exp(c * 1j) >>> d array([ 0.54030231+0.84147098j, -0.84147098+0.54030231j, -0.54030231-0.84147098j]) >>> d.dtype.name 'complex128' Many unary operations, such as computing the sum of all the elements in the array, are implemented as methods of the >>> a = rg.random((2, 3)) >>> a array([[0.82770259, 0.40919914, 0.54959369], [0.02755911, 0.75351311, 0.53814331]]) >>> a.sum() 3.1057109529998157 >>> a.min() 0.027559113243068367 >>> a.max() 0.8277025938204418 By default, these operations apply to the array as though it were a list of numbers, regardless of its shape. However, by specifying the >>> b = np.arange(12).reshape(3, 4) >>> b array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]]) >>> >>> b.sum(axis=0) # sum of each column array([12, 15, 18, 21]) >>> >>> b.min(axis=1) # min of each row array([0, 4, 8]) >>> >>> b.cumsum(axis=1) # cumulative sum along each row array([[ 0, 1, 3, 6], [ 4, 9, 15, 22], [ 8, 17, 27, 38]]) Universal Functions#NumPy provides familiar mathematical functions such as sin, cos, and exp. In NumPy, these are called “universal functions” ( >>> B = np.arange(3) >>> B array([0, 1, 2]) >>> np.exp(B) array([1. , 2.71828183, 7.3890561 ]) >>> np.sqrt(B) array([0. , 1. , 1.41421356]) >>> C = np.array([2., -1., 4.]) >>> np.add(B, C) array([2., 0., 6.]) See also
Indexing, Slicing and Iterating#One-dimensional arrays can be indexed, sliced and iterated over, much like lists and other Python sequences. >>> a = np.arange(10)**3 >>> a array([ 0, 1, 8, 27, 64, 125, 216, 343, 512, 729]) >>> a[2] 8 >>> a[2:5] array([ 8, 27, 64]) >>> # equivalent to a[0:6:2] = 1000; >>> # from start to position 6, exclusive, set every 2nd element to 1000 >>> a[:6:2] = 1000 >>> a array([1000, 1, 1000, 27, 1000, 125, 216, 343, 512, 729]) >>> a[::-1] # reversed a array([ 729, 512, 343, 216, 125, 1000, 27, 1000, 1, 1000]) >>> for i in a: ... print(i**(1 / 3.)) ... 9.999999999999998 1.0 9.999999999999998 3.0 9.999999999999998 4.999999999999999 5.999999999999999 6.999999999999999 7.999999999999999 8.999999999999998 Multidimensional arrays can have one index per axis. These indices are given in a tuple separated by commas: >>> def f(x, y): ... return 10 * x + y ... >>> b = np.fromfunction(f, (5, 4), dtype=int) >>> b array([[ 0, 1, 2, 3], [10, 11, 12, 13], [20, 21, 22, 23], [30, 31, 32, 33], [40, 41, 42, 43]]) >>> b[2, 3] 23 >>> b[0:5, 1] # each row in the second column of b array([ 1, 11, 21, 31, 41]) >>> b[:, 1] # equivalent to the previous example array([ 1, 11, 21, 31, 41]) >>> b[1:3, :] # each column in the second and third row of b array([[10, 11, 12, 13], [20, 21, 22, 23]]) When fewer indices are provided than the number of axes, the missing indices are considered complete slices >>> b[-1] # the last row. Equivalent to b[-1, :] array([40, 41, 42, 43]) The expression within brackets in The
dots (
>>> c = np.array([[[ 0, 1, 2], # a 3D array (two stacked 2D arrays) ... [ 10, 12, 13]], ... [[100, 101, 102], ... [110, 112, 113]]]) >>> c.shape (2, 2, 3) >>> c[1, ...] # same as c[1, :, :] or c[1] array([[100, 101, 102], [110, 112, 113]]) >>> c[..., 2] # same as c[:, :, 2] array([[ 2, 13], [102, 113]]) Iterating over multidimensional arrays is done with respect to the first axis: >>> for row in b: ... print(row) ... [0 1 2 3] [10 11 12 13] [20 21 22 23] [30 31 32 33] [40 41 42 43] However, if one wants to perform an operation on each
element in the array, one can use the >>> for element in b.flat: ... print(element) ... 0 1 2 3 10 11 12 13 20 21 22 23 30 31 32 33 40 41 42 43 Shape Manipulation#Changing the shape of an array#An array has a shape given by the number of elements along each axis: >>> a = np.floor(10 * rg.random((3, 4))) >>> a array([[3., 7., 3., 4.], [1., 4., 2., 2.], [7., 2., 4., 9.]]) >>> a.shape (3, 4) The shape of an array can be changed with various commands. Note that the following three commands all return a modified array, but do not change the original array: >>> a.ravel() # returns the array, flattened array([3., 7., 3., 4., 1., 4., 2., 2., 7., 2., 4., 9.]) >>> a.reshape(6, 2) # returns the array with a modified shape array([[3., 7.], [3., 4.], [1., 4.], [2., 2.], [7., 2.], [4., 9.]]) >>> a.T # returns the array, transposed array([[3., 1., 7.], [7., 4., 2.], [3., 2., 4.], [4., 2., 9.]]) >>> a.T.shape (4, 3) >>> a.shape (3, 4) The order of the elements in the array resulting from The >>> a array([[3., 7., 3., 4.], [1., 4., 2., 2.], [7., 2., 4., 9.]]) >>> a.resize((2, 6)) >>> a array([[3., 7., 3., 4., 1., 4.], [2., 2., 7., 2., 4., 9.]]) If a dimension is given as >>> a.reshape(3, -1) array([[3., 7., 3., 4.], [1., 4., 2., 2.], [7., 2., 4., 9.]]) Stacking together different arrays#Several arrays can be stacked together along different axes: >>> a = np.floor(10 * rg.random((2, 2))) >>> a array([[9., 7.], [5., 2.]]) >>> b = np.floor(10 * rg.random((2, 2))) >>> b array([[1., 9.], [5., 1.]]) >>> np.vstack((a, b)) array([[9., 7.], [5., 2.], [1., 9.], [5., 1.]]) >>> np.hstack((a, b)) array([[9., 7., 1., 9.], [5., 2., 5., 1.]]) The function >>> from numpy import newaxis >>> np.column_stack((a, b)) # with 2D arrays array([[9., 7., 1., 9.], [5., 2., 5., 1.]]) >>> a = np.array([4., 2.]) >>> b = np.array([3., 8.]) >>> np.column_stack((a, b)) # returns a 2D array array([[4., 3.], [2., 8.]]) >>> np.hstack((a, b)) # the result is different array([4., 2., 3., 8.]) >>> a[:, newaxis] # view `a` as a 2D column vector array([[4.], [2.]]) >>> np.column_stack((a[:, newaxis], b[:, newaxis])) array([[4., 3.], [2., 8.]]) >>> np.hstack((a[:, newaxis], b[:, newaxis])) # the result is the same array([[4., 3.], [2., 8.]]) On the other hand, the function >>> np.column_stack is np.hstack False >>> np.row_stack is np.vstack True In general, for arrays with more than two dimensions, Note In complex cases, >>> np.r_[1:4, 0, 4] array([1, 2, 3, 0, 4]) When used with arrays as arguments, Splitting one array into several smaller ones#Using >>> a = np.floor(10 * rg.random((2, 12))) >>> a array([[6., 7., 6., 9., 0., 5., 4., 0., 6., 8., 5., 2.], [8., 5., 5., 7., 1., 8., 6., 7., 1., 8., 1., 0.]]) >>> # Split `a` into 3 >>> np.hsplit(a, 3) [array([[6., 7., 6., 9.], [8., 5., 5., 7.]]), array([[0., 5., 4., 0.], [1., 8., 6., 7.]]), array([[6., 8., 5., 2.], [1., 8., 1., 0.]])] >>> # Split `a` after the third and the fourth column >>> np.hsplit(a, (3, 4)) [array([[6., 7., 6.], [8., 5., 5.]]), array([[9.], [7.]]), array([[0., 5., 4., 0., 6., 8., 5., 2.], [1., 8., 6., 7., 1., 8., 1., 0.]])]
Copies and Views#When operating and manipulating arrays, their data is sometimes copied into a new array and sometimes not. This is often a source of confusion for beginners. There are three cases: No Copy at All#Simple assignments make no copy of objects or their data. >>> a = np.array([[ 0, 1, 2, 3], ... [ 4, 5, 6, 7], ... [ 8, 9, 10, 11]]) >>> b = a # no new object is created >>> b is a # a and b are two names for the same ndarray object True Python passes mutable objects as references, so function calls make no copy. >>> def f(x): ... print(id(x)) ... >>> id(a) # id is a unique identifier of an object 148293216 # may vary >>> f(a) 148293216 # may vary View or Shallow Copy#Different array objects can share the same data. The >>> c = a.view() >>> c is a False >>> c.base is a # c is a view of the data owned by a True >>> c.flags.owndata False >>> >>> c = c.reshape((2, 6)) # a's shape doesn't change >>> a.shape (3, 4) >>> c[0, 4] = 1234 # a's data changes >>> a array([[ 0, 1, 2, 3], [1234, 5, 6, 7], [ 8, 9, 10, 11]]) Slicing an array returns a view of it: >>> s = a[:, 1:3] >>> s[:] = 10 # s[:] is a view of s. Note the difference between s = 10 and s[:] = 10 >>> a array([[ 0, 10, 10, 3], [1234, 10, 10, 7], [ 8, 10, 10, 11]]) Deep Copy#The >>> d = a.copy() # a new array object with new data is created >>> d is a False >>> d.base is a # d doesn't share anything with a False >>> d[0, 0] = 9999 >>> a array([[ 0, 10, 10, 3], [1234, 10, 10, 7], [ 8, 10, 10, 11]]) Sometimes >>> a = np.arange(int(1e8)) >>> b = a[:100].copy() >>> del a # the memory of ``a`` can be released. If Functions and Methods Overview#Here is a list of some useful NumPy functions and methods names ordered in categories. See Routines for the full list. Array Creation
Less Basic#Broadcasting rules#Broadcasting allows universal functions to deal in a meaningful way with inputs that do not have exactly the same shape. The first rule of broadcasting is that if all input arrays do not have the same number of dimensions, a “1” will be repeatedly prepended to the shapes of the smaller arrays until all the arrays have the same number of dimensions. The second rule of broadcasting ensures that arrays with a size of 1 along a particular dimension act as if they had the size of the array with the largest shape along that dimension. The value of the array element is assumed to be the same along that dimension for the “broadcast” array. After application of the broadcasting rules, the sizes of all arrays must match. More details can be found in Broadcasting. Advanced indexing and index tricks#NumPy offers more indexing facilities than regular Python sequences. In addition to indexing by integers and slices, as we saw before, arrays can be indexed by arrays of integers and arrays of booleans. Indexing with Arrays of Indices#>>> a = np.arange(12)**2 # the first 12 square numbers >>> i = np.array([1, 1, 3, 8, 5]) # an array of indices >>> a[i] # the elements of `a` at the positions `i` array([ 1, 1, 9, 64, 25]) >>> >>> j = np.array([[3, 4], [9, 7]]) # a bidimensional array of indices >>> a[j] # the same shape as `j` array([[ 9, 16], [81, 49]]) When the
indexed array >>> palette = np.array([[0, 0, 0], # black ... [255, 0, 0], # red ... [0, 255, 0], # green ... [0, 0, 255], # blue ... [255, 255, 255]]) # white >>> image = np.array([[0, 1, 2, 0], # each value corresponds to a color in the palette ... [0, 3, 4, 0]]) >>> palette[image] # the (2, 4, 3) color image array([[[ 0, 0, 0], [255, 0, 0], [ 0, 255, 0], [ 0, 0, 0]], [[ 0, 0, 0], [ 0, 0, 255], [255, 255, 255], [ 0, 0, 0]]]) We can also give indexes for more than one dimension. The arrays of indices for each dimension must have the same shape. >>> a = np.arange(12).reshape(3, 4) >>> a array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]]) >>> i = np.array([[0, 1], # indices for the first dim of `a` ... [1, 2]]) >>> j = np.array([[2, 1], # indices for the second dim ... [3, 3]]) >>> >>> a[i, j] # i and j must have equal shape array([[ 2, 5], [ 7, 11]]) >>> >>> a[i, 2] array([[ 2, 6], [ 6, 10]]) >>> >>> a[:, j] array([[[ 2, 1], [ 3, 3]], [[ 6, 5], [ 7, 7]], [[10, 9], [11, 11]]]) In Python, >>> l = (i, j) >>> # equivalent to a[i, j] >>> a[l] array([[ 2, 5], [ 7, 11]]) However, we can not do this by putting >>> s = np.array([i, j]) >>> # not what we want >>> a[s] Traceback (most recent call last): File " Another common use of indexing with arrays is the search of the maximum value of time-dependent series: >>> time = np.linspace(20, 145, 5) # time scale >>> data = np.sin(np.arange(20)).reshape(5, 4) # 4 time-dependent series >>> time array([ 20. , 51.25, 82.5 , 113.75, 145. ]) >>> data array([[ 0. , 0.84147098, 0.90929743, 0.14112001], [-0.7568025 , -0.95892427, -0.2794155 , 0.6569866 ], [ 0.98935825, 0.41211849, -0.54402111, -0.99999021], [-0.53657292, 0.42016704, 0.99060736, 0.65028784], [-0.28790332, -0.96139749, -0.75098725, 0.14987721]]) >>> # index of the maxima for each series >>> ind = data.argmax(axis=0) >>> ind array([2, 0, 3, 1]) >>> # times corresponding to the maxima >>> time_max = time[ind] >>> >>> data_max = data[ind, range(data.shape[1])] # => data[ind[0], 0], data[ind[1], 1]... >>> time_max array([ 82.5 , 20. , 113.75, 51.25]) >>> data_max array([0.98935825, 0.84147098, 0.99060736, 0.6569866 ]) >>> np.all(data_max == data.max(axis=0)) True You can also use indexing with arrays as a target to assign to: >>> a = np.arange(5) >>> a array([0, 1, 2, 3, 4]) >>> a[[1, 3, 4]] = 0 >>> a array([0, 0, 2, 0, 0]) However, when the list of indices contains repetitions, the assignment is done several times, leaving behind the last value: >>> a = np.arange(5) >>> a[[0, 0, 2]] = [1, 2, 3] >>> a array([2, 1, 3, 3, 4]) This is reasonable enough, but watch out if you want to use Python’s >>> a = np.arange(5) >>> a[[0, 0, 2]] += 1 >>> a array([1, 1, 3, 3, 4]) Even though 0 occurs twice in the list of indices, the 0th element is only incremented once. This is because Python requires Indexing with Boolean Arrays#When we index arrays with arrays of (integer) indices we are providing the list of indices to pick. With boolean indices the approach is different; we explicitly choose which items in the array we want and which ones we don’t. The most natural way one can think of for boolean indexing is to use boolean arrays that have the same shape as the original array: >>> a = np.arange(12).reshape(3, 4) >>> b = a > 4 >>> b # `b` is a boolean with `a`'s shape array([[False, False, False, False], [False, True, True, True], [ True, True, True, True]]) >>> a[b] # 1d array with the selected elements array([ 5, 6, 7, 8, 9, 10, 11]) This property can be very useful in assignments: >>> a[b] = 0 # All elements of `a` higher than 4 become 0 >>> a array([[0, 1, 2, 3], [4, 0, 0, 0], [0, 0, 0, 0]]) You can look at the following example to see how to use boolean indexing to generate an image of the Mandelbrot set: >>> import numpy as np >>> import matplotlib.pyplot as plt >>> def mandelbrot(h, w, maxit=20, r=2): ... """Returns an image of the Mandelbrot fractal of size (h,w).""" ... x = np.linspace(-2.5, 1.5, 4*h+1) ... y = np.linspace(-1.5, 1.5, 3*w+1) ... A, B = np.meshgrid(x, y) ... C = A + B*1j ... z = np.zeros_like(C) ... divtime = maxit + np.zeros(z.shape, dtype=int) ... ... for i in range(maxit): ... z = z**2 + C ... diverge = abs(z) > r # who is diverging ... div_now = diverge & (divtime == maxit) # who is diverging now ... divtime[div_now] = i # note when ... z[diverge] = r # avoid diverging too much ... ... return divtime >>> plt.clf() >>> plt.imshow(mandelbrot(400, 400)) The second way of indexing with booleans is more similar to integer indexing; for each dimension of the array we give a 1D boolean array selecting the slices we want: >>> a = np.arange(12).reshape(3, 4) >>> b1 = np.array([False, True, True]) # first dim selection >>> b2 = np.array([True, False, True, False]) # second dim selection >>> >>> a[b1, :] # selecting rows array([[ 4, 5, 6, 7], [ 8, 9, 10, 11]]) >>> >>> a[b1] # same thing array([[ 4, 5, 6, 7], [ 8, 9, 10, 11]]) >>> >>> a[:, b2] # selecting columns array([[ 0, 2], [ 4, 6], [ 8, 10]]) >>> >>> a[b1, b2] # a weird thing to do array([ 4, 10]) Note that the length of the 1D boolean array must coincide with the length of the dimension (or axis) you want to slice. In the previous example, The ix_() function#The >>> a = np.array([2, 3, 4, 5]) >>> b = np.array([8, 5, 4]) >>> c = np.array([5, 4, 6, 8, 3]) >>> ax, bx, cx = np.ix_(a, b, c) >>> ax array([[[2]], [[3]], [[4]], [[5]]]) >>> bx array([[[8], [5], [4]]]) >>> cx array([[[5, 4, 6, 8, 3]]]) >>> ax.shape, bx.shape, cx.shape ((4, 1, 1), (1, 3, 1), (1, 1, 5)) >>> result = ax + bx * cx >>> result array([[[42, 34, 50, 66, 26], [27, 22, 32, 42, 17], [22, 18, 26, 34, 14]], [[43, 35, 51, 67, 27], [28, 23, 33, 43, 18], [23, 19, 27, 35, 15]], [[44, 36, 52, 68, 28], [29, 24, 34, 44, 19], [24, 20, 28, 36, 16]], [[45, 37, 53, 69, 29], [30, 25, 35, 45, 20], [25, 21, 29, 37, 17]]]) >>> result[3, 2, 4] 17 >>> a[3] + b[2] * c[4] 17 You could also implement the reduce as follows: >>> def ufunc_reduce(ufct, *vectors): ... vs = np.ix_(*vectors) ... r = ufct.identity ... for v in vs: ... r = ufct(r, v) ... return r and then use it as: >>> ufunc_reduce(np.add, a, b, c) array([[[15, 14, 16, 18, 13], [12, 11, 13, 15, 10], [11, 10, 12, 14, 9]], [[16, 15, 17, 19, 14], [13, 12, 14, 16, 11], [12, 11, 13, 15, 10]], [[17, 16, 18, 20, 15], [14, 13, 15, 17, 12], [13, 12, 14, 16, 11]], [[18, 17, 19, 21, 16], [15, 14, 16, 18, 13], [14, 13, 15, 17, 12]]]) The advantage of this version of reduce compared to the normal ufunc.reduce is that it makes use of the broadcasting rules in order to avoid creating an argument array the size of the output times the number of vectors. Indexing with strings#See Structured arrays. Tricks and Tips#Here we give a list of short and useful tips. “Automatic” Reshaping#To change the dimensions of an array, you can omit one of the sizes which will then be deduced automatically: >>> a = np.arange(30) >>> b = a.reshape((2, -1, 3)) # -1 means "whatever is needed" >>> b.shape (2, 5, 3) >>> b array([[[ 0, 1, 2], [ 3, 4, 5], [ 6, 7, 8], [ 9, 10, 11], [12, 13, 14]], [[15, 16, 17], [18, 19, 20], [21, 22, 23], [24, 25, 26], [27, 28, 29]]]) Vector Stacking#How do we construct a 2D array from a list of equally-sized row vectors? In MATLAB this is quite easy: if >>> x = np.arange(0, 10, 2) >>> y = np.arange(5) >>> m = np.vstack([x, y]) >>> m array([[0, 2, 4, 6, 8], [0, 1, 2, 3, 4]]) >>> xy = np.hstack([x, y]) >>> xy array([0, 2, 4, 6, 8, 0, 1, 2, 3, 4]) The logic behind those functions in more than two dimensions can be strange. Histograms#The NumPy >>> import numpy as np >>> rg = np.random.default_rng(1) >>> import matplotlib.pyplot as plt >>> # Build a vector of 10000 normal deviates with variance 0.5^2 and mean 2 >>> mu, sigma = 2, 0.5 >>> v = rg.normal(mu, sigma, 10000) >>> # Plot a normalized histogram with 50 bins >>> plt.hist(v, bins=50, density=True) # matplotlib version (plot) (array...) >>> # Compute the histogram with numpy and then plot it >>> (n, bins) = np.histogram(v, bins=50, density=True) # NumPy version (no plot) >>> plt.plot(.5 * (bins[1:] + bins[:-1]), n) With Matplotlib >=3.4 you can also use Further reading#
What is the rule for the following sequence of numbers 1 3 7 15?each term is 2⋅immediately preceding term+1.
What is a recurrence relation sequence 1 3 7 15?What is the recurrence relation for the sequence 1, 3, 7, 15, 31, 63,…? Explanation: The recurrence relation for the sequence 1, 3, 7, 15, 31, 63,… should be an = 3an-1−2an-2.
What are the next two numbers in the series 1 3 7 15?1, 3, 7, 15, 31, 63, 127, ....
What is the formula in finding the 5th term in sequence 1 3 7 15?Hence, the 5th term of the sequence is 31.
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