python - 对列表中的条目数进行下采样（无插值）

Question

我有一个包含许多条目的 Python 列表，我需要使用以下任一方法对其进行下采样：

最大行数。例如，将 1234 个条目的列表限制为 1000 个。
原始行的一部分。例如，使列表成为其原始长度的 1/3。

（我需要能够同时使用两种方式，但一次只能使用一种方式）。

我相信对于最大行数，我可以计算所需的比例并将其传递给比例缩小器：

def downsample_to_max(self, rows, max_rows):
        return downsample_to_proportion(rows, max_rows / float(len(rows)))

...所以我真的只需要一个下采样功能。请问有什么提示吗？

编辑：列表包含对象，而不是数值，所以我不需要插值。丢东西就好了。

解决方案：

def downsample_to_proportion(self, rows, proportion):

    counter = 0.0
    last_counter = None
    results = []

    for row in rows:

        counter += proportion

        if int(counter) != last_counter:
            results.append(row)
            last_counter = int(counter)

    return results

谢谢。

score 6 · Accepted Answer

您可以使用islice来自itertools：

from itertools import islice

def downsample_to_proportion(rows, proportion=1):
    return list(islice(rows, 0, len(rows), int(1/proportion)))

用法：

x = range(1,10)
print downsample_to_proportion(x, 0.3)
# [1, 4, 7]

score 3 · Accepted Answer

如果输入已经是序列类型，则直接使用切片语法而不是islice()+list()更有效：

def downsample_to_proportion(rows, proportion):
    return rows[::int(1 / proportion)]

score 2 · Accepted Answer

这个解决方案对于原始海报来说可能有点矫枉过正，但我想我会分享我一直用来解决这个问题和类似问题的代码。

它有点长（大约 90 行），但如果你经常有这个需求，想要一个易于使用的 oneliner，并且需要一个纯 Python 无依赖环境，那么我认为它可能会派上用场。

基本上，您唯一需要做的就是将列表传递给函数并告诉它您希望新列表的长度，该函数将：

如果新长度更小，请通过删除项目来缩小列表，就像之前已经建议的答案一样。
如果新长度较大，则 拉伸/放大您的列表（与缩小尺寸相反），您可以使用添加的选项来决定是否：
- 线性插值 bw 已知值（如果列表包含整数或浮点数，则自动选择）
- 复制每个值，使它们占据新列表的比例大小（如果列表包含非数字，则自动选择）
- 将原始值分开并在两者之间留出空隙

一切都收集在一个函数中，因此如果您需要它，只需将其复制并粘贴到您的脚本中，您就可以立即开始使用它。

例如你可能会说：

origlist = [0,None,None,30,None,50,60,70,None,None,100]
resizedlist = ResizeList(testlist, 21)
print(resizedlist)

并得到

[0, 5.00000000001, 9.9999999999900009, 15.0, 20.000000000010001, 24.999999999989999, 30, 35.0, 40.0, 45.0, 50.0, 55.0, 60.0, 65.0, 70, 75.000000000010004, 79.999999999989996, 85.0, 90.000000000010004, 94.999999999989996, 100]

请注意，由于浮点限制，会出现轻微的不准确性。另外，我是为 Python 2.x 编写的，所以要在 Python 3.x 上使用它，只需添加一行xrange = range.

这是一个在列表列表中定位的子项之间进行插值的绝妙技巧。因此，例如，您可以轻松地在 RGB 颜色元组之间进行插值，以创建 x nr 步长的颜色渐变。假设 RGB 颜色元组列表为 3 和所需的 GRADIENTLENGTH 变量，您可以这样做：

crosssections = zip(*rgbtuples)
grad_crosssections = ( ResizeList(spectrum,GRADIENTLENGTH) for spectrum in crosssections )
rgb_gradient = [list(each) for each in zip(*grad_crosssections)]

它可能需要很多优化，我必须做很多实验。如果您觉得可以改进，请随时编辑我的帖子。这是代码：

def ResizeList(rows, newlength, stretchmethod="not specified", gapvalue=None):
    """
    Resizes (up or down) and returns a new list of a given size, based on an input list.
    - rows: the input list, which can contain any type of value or item (except if using the interpolate stretchmethod which requires floats or ints only)
    - newlength: the new length of the output list (if this is the same as the input list then the original list will be returned immediately)
    - stretchmethod: if the list is being stretched, this decides how to do it. Valid values are:
      - 'interpolate'
        - linearly interpolate between the known values (automatically chosen if list contains ints or floats)
      - 'duplicate'
        - duplicate each value so they occupy a proportional size of the new list (automatically chosen if the list contains non-numbers)
      - 'spread'
        - drags the original values apart and leaves gaps as defined by the gapvalue option
    - gapvalue: a value that will be used as gaps to fill in between the original values when using the 'spread' stretchmethod
    """
    #return input as is if no difference in length
    if newlength == len(rows):
        return rows
    #set auto stretchmode
    if stretchmethod == "not specified":
        if isinstance(rows[0], (int,float)):
            stretchmethod = "interpolate"
        else:
            stretchmethod = "duplicate"
    #reduce newlength 
    newlength -= 1
    #assign first value
    outlist = [rows[0]]
    writinggapsflag = False
    if rows[1] == gapvalue:
        writinggapsflag = True
    relspreadindexgen = (index/float(len(rows)-1) for index in xrange(1,len(rows))) #warning a little hacky by skipping first index cus is assigned auto
    relspreadindex = next(relspreadindexgen)
    spreadflag = False
    gapcount = 0
    for outlistindex in xrange(1, newlength):
        #relative positions
        rel = outlistindex/float(newlength)
        relindex = (len(rows)-1) * rel
        basenr,decimals = str(relindex).split(".")
        relbwindex = float("0."+decimals)
        #determine equivalent value
        if stretchmethod=="interpolate":
            #test for gap
            maybecurrelval = rows[int(relindex)]
            maybenextrelval = rows[int(relindex)+1]
            if maybecurrelval == gapvalue:
                #found gapvalue, so skipping and waiting for valid value to interpolate and add to outlist
                gapcount += 1
                continue
            #test whether to interpolate for previous gaps
            if gapcount > 0:
                #found a valid value after skipping gapvalues so this is where it interpolates all of them from last valid value to this one
                startvalue = outlist[-1]
                endindex = int(relindex)
                endvalue = rows[endindex]
                gapstointerpolate = gapcount 
                allinterpolatedgaps = Resize([startvalue,endvalue],gapstointerpolate+3)
                outlist.extend(allinterpolatedgaps[1:-1])
                gapcount = 0
                writinggapsflag = False
            #interpolate value
            currelval = rows[int(relindex)]
            lookahead = 1
            nextrelval = rows[int(relindex)+lookahead]
            if nextrelval == gapvalue:
                if writinggapsflag:
                    continue
                relbwval = currelval
                writinggapsflag = True
            else:
                relbwval = currelval + (nextrelval - currelval) * relbwindex #basenr pluss interindex percent interpolation of diff to next item
        elif stretchmethod=="duplicate":
            relbwval = rows[int(round(relindex))] #no interpolation possible, so just copy each time
        elif stretchmethod=="spread":
            if rel >= relspreadindex:
                spreadindex = int(len(rows)*relspreadindex)
                relbwval = rows[spreadindex] #spread values further apart so as to leave gaps in between
                relspreadindex = next(relspreadindexgen)
            else:
                relbwval = gapvalue
        #assign each value
        outlist.append(relbwval)
    #assign last value
    if gapcount > 0:
        #this last value also has to interpolate for previous gaps       
        startvalue = outlist[-1]
        endvalue = rows[-1]
        gapstointerpolate = gapcount 
        allinterpolatedgaps = Resize([startvalue,endvalue],gapstointerpolate+3)
        outlist.extend(allinterpolatedgaps[1:-1])
        outlist.append(rows[-1])
        gapcount = 0
        writinggapsflag = False
    else:
        outlist.append(rows[-1])
    return outlist

score 1 · Accepted Answer

1

保留一个计数器，以第二个值递增。每次下限，并在该指数处产生价值。

于 2010-06-10T08:20:11.283 回答

score 1 · Accepted Answer

1

random.choices() 不能解决你的问题吗？更多示例可在此处获得

于 2019-05-07T12:22:15.840 回答

score 0 · Accepted Answer

参考 Ignacio Vazquez-Abrams 的回答：

打印 7 个可用数字中的 3 个：

msg_cache = [1, 2, 3, 4, 5, 6]
msg_n = 3
inc = len(msg_cache) / msg_n
inc_total = 0
for _ in range(0, msg_n):
    msg_downsampled = msg_cache[math.floor(inc_total)]
    print(msg_downsampled)
    inc_total += inc

输出：

0
2
4

对于将许多日志消息下采样到较小的子集很有用。

python - 对列表中的条目数进行下采样（无插值）

6 回答 6

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