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python - python中线性插值的逻辑错误

转载 作者:太空宇宙 更新时间:2023-11-03 20:56:10 25 4
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我的线性插值有一些逻辑错误,它适用于某些情况,但不完全有效。

我尝试使用不同的方式来编写外推情况的逻辑。

def interpolate(x, y, x_test):
for i in range(len(x)):
if x[i] > x_test: #extrapolated condition: when the largest value of
x_below = i - 1 #list x is greater than x_test
x_above = i
y_below = i - 1
y_above = i
break
elif x[i] < x_test: #extrapolated condition: when the largest value of
x_below = i + 1 #list x is greater than x_test
x_above = i
y_below = i + 1
y_above = i
break
else: #interpolated condition: when x_test lies between
return y[i] #two sample points.

#a = (yabove - ybelow) / (xabove - xbelow)
a = (y[y_above] - y[y_below]) / (x[x_above] - x[x_below])
#b = ybelow - a * xbelow
b = y[y_below] - a * x[x_below]
#y’ = a * x’ + b
return a * x_test + b

插值([1, 3, 5], [1, 9, 25], 5.0)我预计输出是 25,但实际输出是 17.0。

最佳答案

我认为您正在寻找这样的东西:

def interpolate(x, y, x_test):
for i in range(len(x)):
if x[i] > x_test: #extrapolated condition: when the largest value of
x_below = i - 1 #list x is greater than x_test
x_above = i
y_below = i - 1
y_above = i
continue # <---- I changed break to continue
elif x[i] < x_test: #extrapolated condition: when the largest value of
x_below = i + 1 #list x is greater than x_test
x_above = i
y_below = i + 1
y_above = i
continue # <---- I changed break to continue
else: #interpolated condition: when x_test lies between
return y[i] #two sample points.

#a = (yabove - ybelow) / (xabove - xbelow)
a = (y[y_above] - y[y_below]) / (x[x_above] - x[x_below])
#b = ybelow - a * xbelow
b = y[y_below] - a * x[x_below]
#y’ = a * x’ + b
return (a * x_test + b)

print(interpolate([1, 3, 5], [1, 9, 25], 5.0))

输出:

25

注意,我将 breaks 更改为 continues

关于python - python中线性插值的逻辑错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56015342/

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