Python’s statistics module is not built for speed, but for precision In the specs for this module, it appears that The built-in sum can lose accuracy when dealing with floats of wildly differing magnitude. Consequently, the above naive mean fails this “torture test” assert mean([1e30, 1, 3, -1e30]) == 1 returning 0 instead of 1, … Read more
You can use colMeans: ### Sample data set.seed(1) m <- data.frame(matrix(sample(100, 20, replace = TRUE), ncol = 4)) ### Your error mean(m) # [1] NA # Warning message: # In mean.default(m) : argument is not numeric or logical: returning NA ### The result using `colMeans` colMeans(m) # X1 X2 X3 X4 # 47.0 64.4 44.8 … Read more
If you don’t specify host while calling app.listen, server will run on all interfaces available i.e on 0.0.0.0 You can bind the IP address using the following code app.listen(3000, ‘127.0.0.1’); If you want to run server in all interface use the following code app.listen(3000, ‘0.0.0.0’); or app.listen(3000)
a.mean() takes an axis argument: In [1]: import numpy as np In [2]: a = np.array([[40, 10], [50, 11]]) In [3]: a.mean(axis=1) # to take the mean of each row Out[3]: array([ 25. , 30.5]) In [4]: a.mean(axis=0) # to take the mean of each col Out[4]: array([ 45. , 10.5]) Or, as a standalone … Read more
The functionality of numpy.mean and tensorflow.reduce_mean are the same. They do the same thing. From the documentation, for numpy and tensorflow, you can see that. Lets look at an example, c = np.array([[3.,4], [5.,6], [6.,7]]) print(np.mean(c,1)) Mean = tf.reduce_mean(c,1) with tf.Session() as sess: result = sess.run(Mean) print(result) Output [ 3.5 5.5 6.5] [ 3.5 5.5 … Read more
I don’t know if Boost has more specific functions, but you can do it with the standard library. Given std::vector<double> v, this is the naive way: #include <numeric> double sum = std::accumulate(v.begin(), v.end(), 0.0); double mean = sum / v.size(); double sq_sum = std::inner_product(v.begin(), v.end(), v.begin(), 0.0); double stdev = std::sqrt(sq_sum / v.size() – mean … Read more
You can use: mse = ((A – B)**2).mean(axis=ax) Or mse = (np.square(A – B)).mean(axis=ax) with ax=0 the average is performed along the row, for each column, returning an array with ax=1 the average is performed along the column, for each row, returning an array with omitting the ax parameter (or setting it to ax=None) the … Read more
If you want to first take mean on the combination of [‘cluster’, ‘org’] and then take mean on cluster groups, you can use: In [59]: (df.groupby([‘cluster’, ‘org’], as_index=False).mean() .groupby(‘cluster’)[‘time’].mean()) Out[59]: cluster 1 15 2 54 3 6 Name: time, dtype: int64 If you want the mean of cluster groups only, then you can use: In … Read more
np.average takes an optional weight parameter. If it is not supplied they are equivalent. Take a look at the source code: Mean, Average np.mean: try: mean = a.mean except AttributeError: return _wrapit(a, ‘mean’, axis, dtype, out) return mean(axis, dtype, out) np.average: … if weights is None : avg = a.mean(axis) scl = avg.dtype.type(a.size/avg.size) else: #code … Read more
I am not aware of anything in the standard library. However, you could use something like: def mean(numbers): return float(sum(numbers)) / max(len(numbers), 1) >>> mean([1,2,3,4]) 2.5 >>> mean([]) 0.0 In numpy, there’s numpy.mean().