平均不等长度的x-y对数据

Question

我有一组稀疏的x-y对数据需要平均。我认为我可以使用一系列搜索索引操作来强制解决问题（因为x数据在重叠时应该匹配），但我觉得应该有一个更好的解决方案，我错过了..

我开始编写代码来生成随机数据，但是在这里显示绘图可能更容易（以对数刻度绘制以突出显示不匹配的数据长度）。

提前致谢。

编辑：这是随机生成数据以匹配感兴趣的数据格式的代码。 ys的平均值是我在保持x位置的同时。

from random import random

# Constructing random example data

x1 = range(1,100)
x2 = range(40,150)
x3 = range(30, 200)
xList = [x1, x2, x3]

y1 = [None for x in x1]
y2 = [None for x in x2]
y3 = [None for x in x3]
yList = [y1, y2 ,y3]

for i in range(0,len(xList)):
    for j in range(0,len(xList[i])):
        yList[i][j] = random()

下面，我将展示我的蛮力方法。我还没有将它概括为for循环，但只是看一看，这就是我要做的。对不起，我不是一个贸易程序员，所以这可能是一个非常迂回的方式。

# Brute force method
minX = min(x1[0], x2[0], x3[0])
maxX = max(x1[-1], x2[-1], x3[-1])
xAll = range(minX, maxX + 1)

x1startIndex = xAll.index(x1[0])
x2startIndex = xAll.index(x2[0])
x3startIndex = xAll.index(x3[0])

x1endIndex = xAll.index(x1[-1])
x2endIndex = xAll.index(x2[-1])
x3endIndex = xAll.index(x3[-1])

from numpy import nan, arange, vstack, nanmean

x1EmptyHead = [nan for x in arange(0,x1startIndex)]  #create empty head
x2EmptyHead = [nan for x in arange(0,x2startIndex)]
x3EmptyHead = [nan for x in arange(0,x3startIndex)]

x1EmptyTail = [nan for x in arange(x1endIndex+1,len(xAll))]  #create empty tail
x2EmptyTail = [nan for x in arange(x2endIndex+1,len(xAll))]
x3EmptyTail = [nan for x in arange(x3endIndex+1,len(xAll))]

y1EqualLength = x1EmptyHead + y1 + x1EmptyTail  #create equal length y-data
y2EqualLength = x2EmptyHead + y2 + x2EmptyTail
y3EqualLength = x3EmptyHead + y3 + x3EmptyTail

yConcat = vstack((y1EqualLength, y2EqualLength, y3EqualLength)) # concatenate
yMean = nanmean(yConcat, axis=0)    # arithmetic mean ignoring NaNs

Answer 1

IIUC，你可以在public static Map<String, String> buildMapFromStringArray( String [] [] stringArray) { if (stringArray == null ) { throw new IllegalArgumentException("buildMapFromStringArray: stringArray is null"); } Map<String, String> map = new HashMap<String, String>( 1 + (2 * stringArray.length) ); for ( String[] keyValue : stringArray) { map.put(keyValue[0], keyValue[1]); } return map; } 中非常干净地完成你所做的事情，因为你可以利用它的索引对齐。例如，如果我从您的示例的较小版本开始：

pandas

给了我

import pandas as pd
df = pd.concat([pd.Series(y,index=x) for x,y in zip(xList, yList)], axis=1)
df["avg"] = df.mean(axis=1)

>>> df 0 1 2 avg 1 0.956034 NaN NaN 0.956034 2 0.947827 NaN NaN 0.947827 3 0.056551 NaN NaN 0.056551 4 0.084872 0.835499 NaN 0.460185 5 NaN 0.735970 NaN 0.735970 6 NaN 0.669730 NaN 0.669730 7 NaN 0.308136 0.581204 0.444670 8 NaN 0.605944 0.158383 0.382164 9 NaN 0.606802 0.430670 0.518736 10 NaN NaN 0.393532 0.393532 11 NaN NaN 0.723012 0.723012 12 NaN NaN 0.994820 0.994820 13 NaN NaN 0.949395 0.949395 14 NaN NaN 0.544177 0.544177 会自动忽略NaN值。然后可以绘制出来：