我写了一个代码,我想用它来减少测量数据。为此,我遍历了 30 组测量数据。在每次迭代中,我使用fsolve求解一组三个非线性方程。这给了我一个包含三个值的数组,然后进一步处理这些值(在下面的示例中lbda,alp, bta, dlt, q, N)。我可以打印结果,但需要在 30 x 6 数组中收集所有 30 个周期的数据,以进行一些统计(即 6 个变量中的每一个的 np.mean)。
在每次迭代结束时,我已经尝试了对我来说最明显的函数np.append, np.vstack,np.concatenates但这只会给我一个 1 x 6 数组,其中仅包含最后一个迭代步骤,而不是包含所有 30 次迭代的所需数组脚步。
# loading data above
m1 = data_arr_blkcorr [:,4] / data_arr_blkcorr [:,2]
m2 = data_arr_blkcorr [:,5] / data_arr_blkcorr [:,2]
m3 = data_arr_blkcorr [:,7] / data_arr_blkcorr [:,2]
N=-1
while (N<29):
N = N+1
T1 = 79.744440299369400
T2 = 4.756431967877120
T3 = 195.146815878103000
T4 = 1.333609171398
T5 = 0.540566631391
T6 = 1
T7 = 1.731261585620
T_all = np.array([T4, T5, T6, T7, T1, T2, T3])
n1 = 0.598169735
n2 = 1.509919737
n3 = 0.600477235
n4 = 0.9364071191658
n5 = 0.5815716133216
n6 = 1
n7 = 1.0455228260642
n_all = np.array([n4, n5, n6, n7, n1, n2, n3])
I1 = 94.905838
I2 = 96.906018
I3 = 97.905405
I4 = 99.907473
I5 = 91.90681
I6 = 93.90509
I7 = 95.90468
# some definition of variables here
A11 = T1-n1
A12 = T2-n2
A13 = T3-n3
A21 = -n1*P1
A22 = -n2*P2
A23 = -n3*P3
A31 = m1[N] * P1
A32 = m2[N] * P2
A33 = m3[N] * P3
b11 = m1[N] - n1
b12 = m2[N] - n2
b13 = m3[N] - n3
# some definition of variables here
T = np.array ([T1, T2, T3])
n = np. array([n1, n2, n3])
m = np.array([m1[N], m2[N], m3[N]])
P = np.array([P1, P2, P3])
def F(x):
return x[0]*T + (1-x[0])*n*np.exp(-x[1]/(1-x[0])*P) - m*np.exp(-x[2]*P)
y = fsolve(F, guess)
lbda = y[0]
alp = y[1]/(1-y[0])
bta = y[2]
dlt = (np.exp(-alp*P2)-1)*1000
N_all = n_all * np.exp(-alp*P_all)
q = (1 + (1 - lbda) / lbda * np.sum(N_all) / np.sum(T_all))**(-1)
print (lbda, alp, bta, dlt, q, N)
浏览帖子我也使用过这个(根据 Koke Cacao 提供的建议):
data_sum = None
new_data = [lbda, alp, bta, dlt, q, N]
data_sum = np.append([data_sum], new_data) if data_sum is not None else new_data
print(data_sum)
胡子哥哥
繁花不似锦
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