array1 = np.array([1,2,3]) scalar =2 # multiply each element in array1 by the scalar valueresult = np.multiply(array1, scalar) print(result) Run Code Output [2 4 6] In this example, we multiplied each element inarray1by the scalar value of2. Example 3: Use out to Store Resul...
modf Return fractional and integral parts of array as a separate array isnan Return boolean array indicating whether each value is NaN (Not a Number) isfinite, isinf Return boolean array indicating whether each element is finite (non-inf, non-NaN) or infinite, respectively cos, cosh, sin, s...
multiply(arr,3) | Multiply each array element by 3 np.divide(arr,4) | Divide each array element by 4 (returns np.nan for division by zero) np.power(arr,5) | Raise each array element to the 5th power #Vector Math#numpy向量计算 np.add(arr1,arr2) | Elementwise add arr2 to arr1...
import numpy as np a = np.array([[1, 2], [3, 4]]) b = np.array([[5, 6], [7, 8]]) path, steps = np.einsum_path('ij,jk->ik', a, b) print(path) print(steps) ['einsum_path', (0, 1)] Complete contraction: ij,jk->ik Naive scaling: 3 Optimized scaling: 3 Naive...
也就是常说的elementwise,需要两个矩阵的大小一样(如果不考虑broadcast的话),multiply函数将两个矩阵相同位置的元素分别相乘,或者直接使用* import numpy as np a = np.array( [ [ 1,2 ], [ 3,4 ] ] ) b = np.array( [ [ 1,2 ], [ 3,4 ] ] ) ...
要创建一个 NumPy 数组,可以使用函数np.array()。 要创建一个简单的数组,您只需向其传递一个列表。如果愿意,还可以指定列表中的数据类型。您可以在这里找到有关数据类型的更多信息。 代码语言:javascript 代码运行次数:0 运行 复制 >>> import numpy as np >>> a = np.array([1, 2, 3]) 您可以通过...
To give you an idea of the performance differnce(性能差异), consider(演示) a NumPy array one million integers, and the equivalent Python list: importnumpyasnp my_arr = np.arange(1000000) my_list =list(range(1000000)) Now let's multiply each sequence by 2: ...
arr[arr<5] | Returns array elements smaller than 5 Scalar Math np.add(arr,1) | Add 1 to each array element np.subtract(arr,2) | Subtract 2 from each array element np.multiply(arr,3) | Multiply each array element by 3 np.divide(arr,4) | Divide each array element by 4 (returns...
v= np.array([1, 0, 1]) vv= np.tile(v, (4, 1))#Stack 4 copies of v on top of each otherprint(vv)#Prints "[[1 0 1]#[1 0 1]#[1 0 1]#[1 0 1]]"y = x + vv#Add x and vv elementwiseprint(y)#Prints "[[ 2 2 4#[ 5 5 7]#[ 8 8 10]#[11 11 13]]"...
(np.array([self.inv_doc_freq[w] for w in words]), (D, 1)) # 计算tfidf矩阵 tfidf = tf * idf # 如果忽略特殊字符 if ignore_special_chars: # 获取特殊字符的索引 idxs = [ self.token2idx["<unk>"], self.token2idx["<eol>"], self.token2idx["<bol>"], ] # 从tfidf矩阵...