我的数据是一组n个观察对及其频率，即每对（xi，yi）对应一些ki，观察次数（xi，yi）。理想情况下，我想计算Kendall的tau和Spearman的rho来计算这些对的所有副本的集合，其中包括k1 + k2 + ... + kn对。问题是观察总数k1 + k2 + ... + kn是巨大的，并且这样的数据结构将不适合存储器。
当然，我考虑分配第i对的频率，ki /（k1 + k2 + ... + kn）作为其权重，并计算加权集的等级相关性 - 但我找不到任何工具那。在我遇到的等级相关的加权变量（例如，scipy.stats.weightedtau）中，权重代表等级的重要性而不是对，这与我的原因无关。 Pearson的r似乎完全具有我需要的加权选项，但它不符合我的目的，因为x和y无关是线性相关的。我想知道我是否错过了关于加权数据点的广义相关的一些概念。
到目前为止我唯一的想法是通过一些公因子c缩小k1，k2，...，kn，因此第i对的缩放份数是[ki / c]（这里[。]是舍入运算符，因为我们需要每对的整数副本）。通过选择c使得[k1 / c] + [k2 / c] + ... + [kn / c]对可以适合存储器，然后我们可以计算得到的集合的相关系数tau和rho。然而，ki和kj可以相差很多个数量级，因此对于某些ki，c可能非常大，因此舍入ki / c会导致信息丢失。
UPD：可以在具有指定频率权重的数据集上计算Spearman的rho和p值，如下所示：
def frequency_pearsonr(data, frequencies):
"""
Calculates Pearson's r between columns (variables), given the
frequencies of the rows (observations).
:param data: 2-D array with data
:param frequencies: 1-D array with frequencies
:return: 2-D array with pairwise correlations,
2-D array with pairwise p-values
"""
df = frequencies.sum() - 2
Sigma = np.cov(data.T, fweights=frequencies)
sigma_diag = Sigma.diagonal()
Sigma_diag_pairwise_products = np.multiply.outer(sigma_diag, sigma_diag)
# Calculate matrix with pairwise correlations.
R = Sigma / np.sqrt(Sigma_diag_pairwise_products)
# Calculate matrix with pairwise t-statistics. Main diagonal should
# get 1 / 0 = inf.
with np.errstate(divide='ignore'):
T = R / np.sqrt((1 - R * R) / df)
# Calculate matrix with pairwise p-values.
P = 2 * stats.t.sf(np.abs(T), df)
return R, P
def frequency_rank(data, frequencies):
"""
Ranks 1-D data array, given the frequency of each value. Same
values get same "averaged" ranks. Array with ranks is shaped to
match the input data array.
:param data: 1-D array with data
:param frequencies: 1-D array with frequencies
:return: 1-D array with ranks
"""
s = 0
ranks = np.empty_like(data)
# Compute rank for each unique value.
for value in sorted(set(data)):
index_grid = np.ix_(data == value)
# Find total frequency of the value.
frequency = frequencies[index_grid].sum()
ranks[index_grid] = s + 0.5 * (frequency + 1)
s += frequency
return ranks
def frequency_spearmanrho(data, frequencies):
"""
Calculates Spearman's rho between columns (variables), given the
frequencies of the rows (observations).
:param data: 2-D array with data
:param frequencies: 1-D array with frequencies
:return: 2-D array with pairwise correlations,
2-D array with pairwise p-values
"""
# Rank the columns.
ranks = np.empty_like(data)
for i, data_column in enumerate(data.T):
ranks[:, i] = frequency_rank(data_column, frequencies)
# Compute Pearson's r correlation and p-values on the ranks.
return frequency_pearsonr(ranks, frequencies)
# Columns are variables and rows are observations, whose frequencies
# are specified.
data_col1 = np.array([1, 0, 1, 0, 1])
data_col2 = np.array([.67, .25, .75, .2, .6])
data_col3 = np.array([.1, .3, .8, .3, .2])
data = np.array([data_col1, data_col2, data_col3]).T
frequencies = np.array([2, 4, 1, 3, 2])
# Same data, but with observations (rows) actually repeated instead of
# their frequencies being specified.
expanded_data_col1 = np.array([1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1])
expanded_data_col2 = np.array([.67, .67, .25, .25, .25, .25, .75, .2, .2, .2, .6, .6])
expanded_data_col3 = np.array([.1, .1, .3, .3, .3, .3, .8, .3, .3, .3, .2, .2])
expanded_data = np.array([expanded_data_col1, expanded_data_col2, expanded_data_col3]).T
# Compute Spearman's rho for data in both formats, and compare.
frequency_Rho, frequency_P = frequency_spearmanrho(data, frequencies)
Rho, P = stats.spearmanr(expanded_data)
print(frequency_Rho - Rho)
print(frequency_P - P)
上面的特定示例显示两种方法都产生相同的相关性和相同的p值：
[[ 0.00000000e+00 0.00000000e+00 0.00000000e+00]
[ 1.11022302e-16 0.00000000e+00 -5.55111512e-17]
[ 0.00000000e+00 -5.55111512e-17 0.00000000e+00]]
[[ 0.00000000e+00 -1.35525272e-19 4.16333634e-17]
[ -9.21571847e-19 0.00000000e+00 -5.55111512e-17]
[ 4.16333634e-17 -5.55111512e-17 0.00000000e+00]]
保罗建议的计算Kendall's tau的方法很有效。您不必将已排序数组的索引指定为排名，未排序的索引也可以同样正常工作（如加权tau示例中所示）。权重也不需要标准化。
常规（未加权）Kendall的tau（在“扩展”数据集上）：
stats.kendalltau([0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1],
[.25, .25, .25, .25, .2, .2, .2, .667, .667, .75, .6, .6])
KendalltauResult(correlation=0.7977240352174656, pvalue=0.0034446936330652677)
加权Kendall的tau（在出现次数为权重的数据集上）：
stats.weightedtau([1, 0, 1, 0, 1],
[.667, .25, .75, .2, .6],
rank=False,
weigher=lambda r: [2, 4, 1, 3, 2][r],
additive=False)
WeightedTauResult(correlation=0.7977240352174656, pvalue=nan)
现在，由于weightedtau实现的特殊性，p值永远不会被计算出来。我们可以用最初提供的缩减事件的技巧来近似p值，但我非常欣赏其他方法。基于可用内存量的算法行为决策看起来像是痛苦。