np.isnan(x)#判断目标x是否包含NaNnp.where(np.isnan(x))#判断x中所含有的nan数据所在的位置np.nan_to_num(x)#replace nan with zero and inf with finite numbers 当使用 np.isnan(x) 函数后,由于数组内容过多,导致无法查询到是否存在某个数组元素为NaN 通过np.where来查询x数
出现Nan值的情况，总的来说，TensorFlow中出现Nan值的情况有两种，一种是在loss中计算后得到了Nan值，...
returns index positionnp.where(y>5)array([2, 3, 5, 7, 8], dtype=int64),)# First will replace the values that match the condition,# second will replace the values that does notnp.where(y>5, "Hit", "Miss")array(['Miss', 'Miss', 'Hit...
numpy.isnan(element) Note: 不能使用array[0] == np.NaN,总是返回False! numpy数组元素替换numpy.nan_to_num(x) 判断某元素是否是nan,inf,neginf,如果是,nan换为0,inf换为一个非常大的数,neginf换为非常小的数 numpy.nan_to_num(x)Replace nan with zero and inf with finite numbers.Returns an ...
(3).np.nan_to_num:Replace nan with zero and inf with finite numbers. 把np.nan(非常小的数字,接近0)用0取代 np.inf,或者-np.inf(正负无穷大的数字)用有限数替代 np.set_printoptions(precision=8) x = np.array([np.inf, -np.inf, np.nan, -128, 128]) ...
#zeros 函数生成元素全部为0的数组, print(np.zeros((3,4))) #empty函数生成元素没有赋值的数组,这时元素值由内存中原来的内容决定。 print(np.empty((2,3))) #ones函数生成元素全部为1的数组 print(np.ones((2,3,4),dtype = np.int16)) ...
array([2, 3, 5, 7, 8], dtype=int64) # First will replace the values that match the condition, # second will replace the values that does not >>> np.where(y>5, "Hit", "Miss") array(['Miss', 'Miss', 'Hit', 'Hit', 'Miss', ...
y = np.array([1,5,6,8,1,7,3,6,9])# Where y is greater than 5, returns index positionnp.where(y>5)array([2, 3, 5, 7, 8], dtype=int64),)# First will replace the values that match the condition, # second will replace the values that does notnp.where(y>5, "Hit", "...
Suppose that we are given a NumPy array that contains some NaN values and we need to replace these NaN values with the closest numerical value (non-NaN values).How to replace NaN's in NumPy array with closest non-NaN value?To replace NaN's in NumPy array with closest non-NaN value, ...
0 * np.nannp.nan == np.nan np.inf > np.nannp.nan - np.nan0.3 == 3 * 0.1 16、创建一个5 × 5矩阵，对角线值为1,2,3,4 Z = np.diag(1+np.arange(4),k=-1)print(Z)17、创建一个8x8的矩阵，并使用0，1间隔填充 Z = np.zeros((8,8),dtype=int)Z[1::2,::2] = 1Z[:...