在python函数中拆分数据时保持比率

Question

我有一些数据，我想将其分成保持共同比率的较小组。我写了一个函数，它将接受两个数组的输入并计算大小比，然后告诉我可以将它分成多少组的选项（如果所有组的大小相同），这里是函数：

def cross_validation_group(train_data, test_data):
    import numpy as np
    from calculator import factors
    test_length = len(test_data)
    train_length = len(train_data)
    total_length = test_length + train_length
    ratio = test_length/float(total_length)
    possibilities = factors(total_length)
    print possibilities
    print possibilities[len(possibilities)-1] * ratio
    super_count = 0
    for i in possibilities:
        if i < len(possibilities)/2:
            pass
        else: 
            attempt = float(i * ratio)
            if attempt.is_integer():
                print str(i) + " is an option for total size with " +  str(attempt) + " as test size and " + str(i - attempt) + " as train size! This is with " + str(total_length/i) + " folds."
            else:
                pass
    folds = int(raw_input("So how many folds would you like to use? If no possibilities were given that would be sufficient, type 0: "))
    if folds != 0:
        total_size = total_length/folds
        test_size = float(total_size * ratio)
        train_size = total_size - test_size
        columns = train_data[0]
        columns= len(columns)
        groups = np.empty((folds,(test_size + train_size),columns))
        i = 0
        a = 0
        b = 0
        for j in range (0,folds):
            test_size_new = test_size * (j + 1)
            train_size_new = train_size * j
            total_size_new = (train_size + test_size) * (j + 1)
            cut_off = total_size_new - train_size
            p = 0
            while i < total_size_new:
                if i < cut_off:
                    groups[j,p] = test_data[a]
                    a += 1
                else:
                    groups[j,p] = train_data[b]
                    b += 1
                i += 1
                p += 1
        return groups
    else:
        print "This method cannot be used because the ratio cannot be maintained with equal group sizes other than for the options you were givens"

所以我的问题是我如何才能使函数的第三个输入成为折叠数并更改函数，而不是迭代以确保每个组具有相同的数量正确的比例，它只会有正确的比例，但大小不一？

@JamesHolderness 的补充

所以你的方法几乎是完美的，但这里有一个问题：

长度为 357 和 143，9 折，这是返回列表：

[(39, 16), (39, 16), (39, 16), (39, 16), (39, 16), (39, 16), (39, 16), (39, 16), (39, 16)]

现在，当您将列相加时，您会得到：351 144

351 很好，因为它小于 357，但 144 不起作用，因为它大于 143！原因是357和143是数组的长度，所以那个数组的第144行不存在……

Answer 1

这是一个我认为可能对你有用的算法。

您将 test_length 和 train_length 除以它们的 GCD，得到一个简单分数的比率。你把分子和分母加在一起，这就是你的组的大小因素。

例如如果比例是 3:2，那么每组的大小必须是 5 的倍数。

然后，您将总长度除以折叠数，以获得第一组的理想大小，这很可能是一个浮点数。您找到小于或等于 5 的最大倍数，这就是您的第一组。

从您的总数中减去该值，然后除以 folds-1 以获得下一组的理想尺寸。再次找到 5 的最大倍数，从总数中减去 ，然后继续，直到计算完所有组。

一些示例代码：

total_length = test_length + train_length          
divisor = gcd(test_length,train_length)
test_multiple = test_length/divisor
train_multiple = train_length/divisor
total_multiple = test_multiple + train_multiple 

# Adjust the ratio if there isn't enough data for the requested folds
if total_length/total_multiple < folds:
  total_multiple = total_length/folds
  test_multiple = int(round(float(test_length)*total_multiple/total_length))
  train_multiple = total_multiple - test_multiple

groups = []
for i in range(folds,0,-1):
  float_size = float(total_length)/i
  int_size = int(float_size/total_multiple)*total_multiple
  test_size = int_size*test_multiple/total_multiple
  train_size = int_size*train_multiple/total_multiple
  test_length -= test_size    # keep track of the test data used
  train_length -= train_size  # keep track of the train data used
  total_length -= int_size
  groups.append((test_size,train_size))

# If the test_length or train_length are negative, we need to adjust the groups
# to "give back" some of the data.
distribute_overrun(groups,test_length,0)
distribute_overrun(groups,train_length,1)

这已更新，以跟踪每个组（测试和训练）使用的大小，但如果我们最初使用过多也不必担心。

最后，如果有任何超支（即test_length 或train_length 变为负数），我们会通过在尽可能多的项目中减少比率的适当一侧来将超支分配回组中使溢出恢复为零。

distribute_overrun 函数包含在下面。

def distribute_overrun(groups,overrun,part):
    i = 0
    while overrun < 0:
      group = list(groups[i])
      group[part] -= 1
      groups[i] = tuple(group)
      overrun += 1
      i += 1

最后，groups 将是一个包含每个组的 test_size 和 train_size 的元组列表。

如果这听起来像你想要的，但你需要我扩展代码示例，请告诉我。

Answer 2

在另一个问题中，作者想进行与您类似的交叉验证。请take a look to this answer。找出问题的答案，就像：

import numpy as np
# in both train_data the first line is used for the cross-validation,
# and the other lines will follow, so you can add as many lines as you want
test_data = np.array([ 0.,  1.,  2.,  3.,  4.,  5.])
train_data  = np.array([[ 0.09,  1.9,  1.1,  1.5,  4.2,  3.1,  5.1],
                       [    3,    4,  3.1,   10,   20,    2,    3]])

def cross_validation_group( test_data, train_data):
    om1,om2 = np.meshgrid(test_data,train_data[0])
    dist = (om1-om2)**2
    indexes = np.argsort( dist, axis=0 )
    return train_data[:, indexes[0]]

print cross_validation_group( test_data, train_data )
# array([[  0.09,   1.1 ,   1.9 ,   3.1 ,   4.2 ,   5.1 ],
#        [     3 ,  3.1 ,     4 ,     2 ,    20 ,     3 ]])

您将拥有与test_data 中定义的间隔相对应的train_data。