我需要对X数组中的值对应的Y值求平均值...
X=np.array([ 1, 1, 2, 2, 2, 2, 3, 3 ... ])
Y=np.array([ 10, 30, 15, 10, 16, 10, 15, 20 ... ])
换句话说,X数组中1值的相当于Y数组中的10和30,它们的平均值为20,2值的相当于15、10、16和10,其平均值为12.75,以此类推...
如何计算这些平均值?
其中一种选择是使用线性回归的一个属性(与分类变量相关):
import numpy as np
x = np.array([ 1, 1, 2, 2, 2, 2, 3, 3 ])
y = np.array([ 10, 30, 15, 10, 16, 10, 15, 20 ])
x_dummies = x[:, None] == np.unique(x)
means = np.linalg.lstsq(x_dummies, y, rcond=None)[0]
print(means) # [20. 12.75 17.5 ]
pandas
。import pandas as pd
import numpy as np
N = pd.DataFrame(np.transpose([X,Y]),
columns=['X', 'Y']).groupby('X')['Y'].mean().to_numpy()
# array([20. , 12.75, 17.5 ])
pd.Series(Y).groupby(X).mean().to_numpy()
;) - mozwayimport numpy as np
X = np.array([ 1, 1, 2, 2, 2, 2, 3, 3])
Y = np.array([ 10, 30, 15, 10, 16, 10, 15, 20])
# Only unique values
unique_vals = np.unique(X);
# Loop for every value
for val in unique_vals:
# Search for proper indexes in Y
idx = np.where(X == val)
# Mean for finded indexes
aver = np.mean(Y[idx])
print(f"Average for {val}: {aver}")
结果:
1的平均值:20.0
2的平均值:12.75
3的平均值:17.5
你可以像下面这样使用:
import numpy as np
X=np.array([ 1, 1, 2, 2, 2, 2, 3, 3])
Y=np.array([ 10, 30, 15, 10, 16, 10, 15, 20])
def groupby(a, b):
# Get argsort indices, to be used to sort a and b in the next steps
sidx = b.argsort(kind='mergesort')
a_sorted = a[sidx]
b_sorted = b[sidx]
# Get the group limit indices (start, stop of groups)
cut_idx = np.flatnonzero(np.r_[True,b_sorted[1:] != b_sorted[:-1],True])
# Split input array with those start, stop ones
out = [a_sorted[i:j] for i,j in zip(cut_idx[:-1],cut_idx[1:])]
return out
group_by_array=groupby(Y,X)
for item in group_by_array:
print(np.average(item))
我认为这个解决方案应该可行:
avg_arr = []
i = 1
while i <= np.max(x):
inds = np.where(x == i)
my_val = np.average(y[inds[0][0]:inds[0][-1]])
avg_arr.append(my_val)
i+=1
虽然不是最干净的,但我能够快速测试它,而且它确实有效。
np.bincount(X-1, Y) / np.bincount(X-1)
。 - Michael Szczesny