numpy

问题 A NaN of type decimal.Decimal causes: math.isnan to return True numpy.isnan to throw a TypeError exception. pandas.isnull to return False What is the difference between math.isnan, numpy.isnan and pandas.isnull? 回答1: The only difference between math.isnan and numpy.isnan is that numpy.isnan can handle lists, arrays, tuples whereas math.isnan can ONLY handle single integers or floats. However , I suggest using math.isnan when you just want to check if a number is nan because numpy takes

问题 Suppose A = np.array([[1,2,0,3,5,0,0],[5,6,7,0,9,5,10]]) (The actual data set in this case is a square matrix with a shifted diagonal of zeros) I want to create an np array V that sums the pairs reflected across the first zero in each row, in the order of their distance from the zero. i.e., V[0]=[(2+3),(1+5),0] V[1] =[(7+9),(6+5),(10+5)]... So V = ([5,6,0],[16,11,15]) I can accomplish this in a very rudimentary way by looping through each row and then adding each row to a shifted version of

问题 For example, I want to add the 2nd element of this tensor where the 1st element is same. Any Numpy based solution is also welcomed! From : x = tf.constant([ [1., 0.9], [2., 0.7], [1., 0.7], [3., 0.4], [4., 0.8] ], dtype=tf.float32) To: x = tf.constant([ [1., 1.6], [2., 0.7], [3., 0.4], [4., 0.8] ], dtype=tf.float32) 回答1: numpy solution: x = np.array([ [1., 0.9], [2., 0.7], [1., 0.7], [3., 0.4], [4., 0.8]]) ans = np.array([[i,np.sum(x[np.where(x[:,0]==i), 1])] for i in set(x[:,0])]) gives

问题 For example, I want to add the 2nd element of this tensor where the 1st element is same. Any Numpy based solution is also welcomed! From : x = tf.constant([ [1., 0.9], [2., 0.7], [1., 0.7], [3., 0.4], [4., 0.8] ], dtype=tf.float32) To: x = tf.constant([ [1., 1.6], [2., 0.7], [3., 0.4], [4., 0.8] ], dtype=tf.float32) 回答1: numpy solution: x = np.array([ [1., 0.9], [2., 0.7], [1., 0.7], [3., 0.4], [4., 0.8]]) ans = np.array([[i,np.sum(x[np.where(x[:,0]==i), 1])] for i in set(x[:,0])]) gives

问题 So below is the code I have been using. I'm a bit of a newb. I've been testing with just the head of the data because of the quota for using the API. Below is a snapshot of the dataframe: latitude longitude 0 -73.99107 40.730054 1 -74.000193 40.718803 2 -73.983849 40.761728 3 -73.97499915 40.68086214 4 -73.89488591 40.66471445 This is where I am getting tripped up. train['latlng'] = train.apply(lambda row: '{},{}'.format(row['latitude'], row['longitude']), axis=1) train['geocode_data'] =

问题 I have a volume of image slices and their according masks. I've been trying to use skimage.segmentation library to mark the object in mind for each slice according to its mask. import numpy as np from skimage.segmentation import mark_boundaries import matplotlib.pyplot as plt def plot_marked_volume(marked_image_volume, mask): for slice in range(len(marked_image_volume)): if np.count_nonzero(mask[slice,:,:]): plt.figure(figsize=(10,10)) edges_pz = mark_boundaries(marked_image_volume[slice,:,:]

问题 So I have around 13 years of stock market data of daily low high open close. The problem is the markets are closed sometimes in between and hence Monday to Friday might not appear continuously sometimes. Look below Date Day Open High Low Close Adjusted Close 0 17-09-2007 Monday 6898 6977.2 6843 6897.1 6897.100098 1 18-09-2007 Tuesday 6921.15 7078.95 6883.6 7059.65 7059.649902 2 19-09-2007 Wednesday 7111 7419.35 7111 7401.85 7401.850098 3 20-09-2007 Thursday 7404.95 7462.9 7343.6 7390.15 7390

问题 I am using the following code to calculate the quartiles of a given data set: #!/usr/bin/python import numpy as np series = [1,2,2,2,2,2,2,2,2,2,2,5,5,6,7,8] p1 = 25 p2 = 50 p3 = 75 q1 = np.percentile(series, p1) q2 = np.percentile(series, p2) q3 = np.percentile(series, p3) print('percentile(' + str(p1) + '): ' + str(q1)) print('percentile(' + str(p2) + '): ' + str(q2)) print('percentile(' + str(p3) + '): ' + str(q3)) The percentile function returns the quartiles, however, I would also like

问题 I am using the following code to calculate the quartiles of a given data set: #!/usr/bin/python import numpy as np series = [1,2,2,2,2,2,2,2,2,2,2,5,5,6,7,8] p1 = 25 p2 = 50 p3 = 75 q1 = np.percentile(series, p1) q2 = np.percentile(series, p2) q3 = np.percentile(series, p3) print('percentile(' + str(p1) + '): ' + str(q1)) print('percentile(' + str(p2) + '): ' + str(q2)) print('percentile(' + str(p3) + '): ' + str(q3)) The percentile function returns the quartiles, however, I would also like

问题 So I have around 13 years of stock market data of daily low high open close. The problem is the markets are closed sometimes in between and hence Monday to Friday might not appear continuously sometimes. Look below Date Day Open High Low Close Adjusted Close 0 17-09-2007 Monday 6898 6977.2 6843 6897.1 6897.100098 1 18-09-2007 Tuesday 6921.15 7078.95 6883.6 7059.65 7059.649902 2 19-09-2007 Wednesday 7111 7419.35 7111 7401.85 7401.850098 3 20-09-2007 Thursday 7404.95 7462.9 7343.6 7390.15 7390