Question

我有两个我想重新调整大小的数组，但我也希望保留原始值。下面的代码重新调整了数组的大小，但问题是它覆盖了原始值，正如您在查看

的输出时所看到的那样

print(x) 
print(y)

脚本末尾的

命令。但是，如果我们注释掉这一行

# NewX,NewY=resize(x,y,xmin=MinRR,xmax=MaxRR,ymin=minLVET,ymax=maxLVET)

然后x和y的原始值正确打印出来。但是，如果我们删除注释并保留代码，那么x和y显然会被覆盖，因为

print(x) 
print(y)

命令然后分别输出NewX和NewY的值。

我的代码如下。 任何人都可以告诉我如何修复下面的代码，以便x和y保留其原始值，以便NewX和NewY获得新调整大小的值吗？

import numpy as np

def GetMinRR(age):
    MaxHR = 208-(0.7*age)
    MinRR = (60/MaxHR)*1000
    return MinRR

def resize(x,y,xmin=0.0,xmax=1.0,ymin=0.0,ymax=1.0):
    # Create local variables
    NewX = x
    NewY = y
    # If the mins are greater than the maxs, then flip them.
    if xmin>xmax: xmin,xmax=xmax,xmin 
    if ymin>ymax: ymin,ymax=ymax,ymin
    #----------------------------------------------------------------------------------------------    
    # The rest of the code below re-calculates all the values in x and then in y with these steps:
    #       1.) Subtract the actual minimum of the input x-vector from each value of x
    #       2.) Multiply each resulting value of x by the result of dividing the difference
    #           between the new xmin and xmax by the actual maximum of the input x-vector
    #       3.) Add the new minimum to each value of x
    # Note: I wrote in x-notation, but the identical process is also repeated for y
    #----------------------------------------------------------------------------------------------    
    # Subtracts right operand from the left operand and assigns the result to the left operand.
    # Note: c -= a is equivalent to c = c - a
    NewX -= x.min()

    # Multiplies right operand with the left operand and assigns the result to the left operand.
    # Note: c *= a is equivalent to c = c * a
    NewX *= (xmax-xmin)/NewX.max()

    # Adds right operand to the left operand and assigns the result to the left operand.
    # Note: c += a is equivalent to c = c + a
    NewX += xmin

    # Subtracts right operand from the left operand and assigns the result to the left operand.
    # Note: c -= a is equivalent to c = c - a
    NewY -= y.min()

    # Multiplies right operand with the left operand and assigns the result to the left operand.
    # Note: c *= a is equivalent to c = c * a
    NewY *= (ymax-ymin)/NewY.max()

    # Adds right operand to the left operand and assigns the result to the left operand.
    # Note: c += a is equivalent to c = c + a
    NewY += ymin

    return (NewX,NewY)

# Declare raw data for use in creating logistic regression equation
x = np.array([821,576,473,377,326],dtype='float') 
y = np.array([255,235,208,166,157],dtype='float') 

# Call resize() function to re-calculate coordinates that will be used for equation
MinRR=GetMinRR(34)
MaxRR=1200
minLVET=(y[4]/x[4])*MinRR
maxLVET=(y[0]/x[0])*MaxRR
NewX,NewY=resize(x,y,xmin=MinRR,xmax=MaxRR,ymin=minLVET,ymax=maxLVET) 

print 'x is:  ',x 
print 'y is:  ',y

Answer 1

NewX = x.copy()
NewY = y.copy()

numpy数组也支持__copy__接口，可以使用复制模块进行复制，因此也可以使用：

NewX = copy.copy(x)
NewY = copy.copy(y)

如果您希望按原样保留函数的当前行为，则需要将x和y的所有出现替换为NewX和NewY 。如果函数的当前行为是错误的，您可以按原样保留它们。

Answer 2

在x中明确复制y和resize：

def resize(...):
    NewX = [t for t in x]
    NewY = [t for t in y]

Python总是通过引用传递，因此您在子例程中所做的任何更改都是对实际传递的对象进行的。

Answer 3

原始resize重演。为x完成的所有事情都会重复y。 That's not good，因为这意味着您必须维护两倍于您真正需要的代码。解决方案是让resize只对一个数组起作用，然后调用它两次（或根据需要）：

def resize(arr,lower=0.0,upper=1.0):
    # Create local variables
    result = arr.copy()
    # If the mins are greater than the maxs, then flip them.
    if lower>upper: lower,upper=upper,lower 
    #----------------------------------------------------------------------------------------------    
    # The rest of the code below re-calculates all the values in x and then in y with these steps:
    #       1.) Subtract the actual minimum of the input x-vector from each value of x
    #       2.) Multiply each resulting value of x by the result of dividing the difference
    #           between the new lower and upper by the actual maximum of the input x-vector
    #       3.) Add the new minimum to each value of x
    # Note: I wrote in x-notation, but the identical process is also repeated for y
    #----------------------------------------------------------------------------------------------    
    # Subtracts right operand from the left operand and assigns the result to the left operand.
    # Note: c -= a is equivalent to c = c - a
    result -= result.min()

    # Multiplies right operand with the left operand and assigns the result to the left operand.
    # Note: c *= a is equivalent to c = c * a
    result *= (upper-lower)/result.max()

    # Adds right operand to the left operand and assigns the result to the left operand.
    # Note: c += a is equivalent to c = c + a
    result += lower
    return result

这样称呼：

NewX=resize(x,lower=MinRR,upper=MaxRR)
NewY=resize(y,lower=minLVET,upper=maxLVET)

在numpy数组中重新调整数据大小时出错

3 个答案: