如何使用 Numpy.polyfit 绘制趋势

Question

感谢用户 Eduard Ilyasov 几天前帮助我

现在我得到了一些结果，但我很难理解这些

我正在尝试计算从 1979 年到 2016 年的温度趋势。

    #calculate trend
    ####numpy.ployfit
    nmon = nyr * 12
    tdum = MA.arange(0,nmon)
    ntimes, ny, nx = tempF.shape 
#ntimes is time, ny is latitude, nx is longitude 
print tempF.shape

trend = MA.zeros((ny,nx), dtype='2f')
#trend = MA.zeros((ny,nx),dtype=float)

print trend.shape

for y in range (0,ny):
    for x in range (0,nx):
        trend[y,x]= numpy.polyfit(tdum, tempF[:,y,x],1)

print trend.shape
print trend

这些是结果：

(
(456, 241, 480)
(241, 480, 2)
(241, 480, 2)
[[[ 0.00854342 -1.94362879]
  [ 0.00854342 -1.94362879]
  [ 0.00854342 -1.94362879]
  ..., 
  [ 0.00854342 -1.94362879]
  [ 0.00854342 -1.94362879]
  [ 0.00854342 -1.94362879]]

 [[ 0.00824162 -1.87496781]
  [ 0.00824792 -1.87640166]
  [ 0.00825524 -1.87806702]
  ..., 
  [ 0.00822667 -1.87156749]
  [ 0.00823172 -1.87271607]
  [ 0.0082366  -1.87382615]]

 [[ 0.00767854 -1.7468679 ]
  [ 0.00769076 -1.74964726]
  [ 0.00770384 -1.75262356]
  ..., 
  [ 0.00764879 -1.74010038]
  [ 0.00765911 -1.74244869]
  [ 0.00766829 -1.74453557]]

 ..., 
 [[-0.0025295   0.57546186]
  [-0.00252633  0.57474071]
  [-0.00252274  0.57392275]
  ..., 
  [-0.00253488  0.57668549]
  [-0.00253269  0.57618785]
  [-0.00253125  0.57585901]]

 [[-0.00315533  0.71783835]
  [-0.00315261  0.71721852]
  [-0.00314936  0.71648043]
  ..., 
  [-0.00315671  0.71815109]
  [-0.00315621  0.71803892]
  [-0.00315584  0.71795386]]

 [[-0.00309109  0.7032221 ]
  [-0.00309109  0.7032221 ]
  [-0.00309109  0.7032221 ]
  ..., 
  [-0.00309109  0.7032221 ]
  [-0.00309109  0.7032221 ]
  [-0.00309109  0.7032221 ]]]

我理解每个括号中的第二个值应该是趋势值的系数，但我不了解趋势的形状。每个 [] 中第一个数字的含义是什么？我应该使用什么趋势值来绘制趋势图？

如果你能帮助我，非常感谢

Answer 1

如果您进一步阅读 numpy.polyfit() 的文档，您将看到此函数的定义

The solution minimizes the squared error

E = \sum_{j=0}^k |p(x_j) - y_j|^2

in the equations:
x[0]**n * p[0] + ... + x[0] * p[n-1] + p[n] = y[0]
x[1]**n * p[0] + ... + x[1] * p[n-1] + p[n] = y[1]
...
x[k]**n * p[0] + ... + x[k] * p[n-1] + p[n] = y[k]

对于趋势是线性的情况，这意味着 trend[y,x,0] 是 trend 的值（也称为 slope) 和 trend[y,x,1] 是截距 .

为了便于说明，请考虑以下示例：

import numpy as np
from matplotlib import pyplot as plt
N = 10
xs = np.random.random(N)
ys = np.random.random(N)
trend = np.polyfit(xs,ys,1)
plt.plot(xs,ys,'o')
trendpoly = np.poly1d(trend) 
plt.plot(xs,trendpoly(xs))

如何使用 Numpy.polyfit 绘制趋势

how to use Numpy.polyfit to plot trend

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