查找忽略 NA 的数据帧之间的协方差

Question

我有一个数据框，其中每一列都是一个时间序列。我想找到其中一列与数据框其余部分之间的相关性。问题是，数据帧的其余部分包含 NA。有没有一种简洁的方法可以找到所有这些忽略每列中的 NA 的协方差？

我只能找到一种使用 for 循环的方法，这并不理想。 这是一个示例数据

structure(list(Date = structure(c(18628, 18629, 18630, 18631, 
18632, 18633, 18634, 18635, 18636, 18637), class = "Date"), X1 = c(NA, 
NA, NA, NA, 1.16092168555067, 0.591202293337843, -0.279052669225263, 
-0.780435476613128, -0.852870619718068, -0.708611614262357), 
    X2 = c(NA, NA, -0.222767493777229, 1.50328295132467, 0.934670132217215, 
    1.37678188537077, 0.343280062984192, 1.23279081824003, -1.08074586121729, 
    0.208120194894818), X3 = c(NA, NA, NA, NA, NA, 1.72057538716556, 
    1.37803710718683, 1.24717457500191, -0.00930256437131184, 
    0.491423553538728), X4 = c(1.15304498847709, -0.154433520961086, 
    -0.361871232243227, -0.981985961481073, 0.596667113671836, 
    -0.0960746707238904, -1.53792603627306, 1.00296956396233, 
    0.128292175597246, -1.12744557711187)), row.names = c(NA, 
-10L), class = "data.frame")

使用这些数据，我想获得一个向量，它具有 x1 和 x4 之间的协方差，删除 x1 中的所有 NA，然后是 x2 和 x4 之间的协方差，删除 x2 中的所有 NA，等等。我

Answer 1

我认为您可以在函数cov() 中使用参数use = complete 来忽略NAs 值（也可以使用其他选项）。这可能对您有用（该示例使用您的数据框计算 var-cov 矩阵。

aa = structure(list(Date = structure(c(18628, 18629, 18630, 18631, 
18632, 18633, 18634, 18635, 18636, 18637), class = "Date"), X1 = c(NA, 
NA, NA, NA, 1.16092168555067, 0.591202293337843, -0.279052669225263, 
-0.780435476613128, -0.852870619718068, -0.708611614262357), 
    X2 = c(NA, NA, -0.222767493777229, 1.50328295132467, 0.934670132217215, 
    1.37678188537077, 0.343280062984192, 1.23279081824003, -1.08074586121729, 
    0.208120194894818), X3 = c(NA, NA, NA, NA, NA, 1.72057538716556, 
    1.37803710718683, 1.24717457500191, -0.00930256437131184, 
    0.491423553538728), X4 = c(1.15304498847709, -0.154433520961086, 
    -0.361871232243227, -0.981985961481073, 0.596667113671836, 
    -0.0960746707238904, -1.53792603627306, 1.00296956396233, 
    0.128292175597246, -1.12744557711187)), row.names = c(NA, 
-10L), class = "data.frame")

cov(aa[, -1], use = 'complete')

            X1        X2           X3           X4
X1  0.36049927 0.3436964  0.319734021 -0.095666285
X2  0.34369636 0.9697499  0.620778723  0.220293460
X3  0.31973402 0.6207787  0.498663689 -0.003728815
X4 -0.09566628 0.2202935 -0.003728815  1.034121716

编辑成对完整的 obs use = 'pairwise.complete.obs'

cov(aa[, -1], use = 'pairwise.complete.obs')

          X1         X2           X3           X4
X1 0.6975825 0.41039379  0.319734021  0.164427330
X2 0.4103938 0.80303434  0.620778723  0.091645141
X3 0.3197340 0.62077872  0.498663689 -0.003728815
X4 0.1644273 0.09164514 -0.003728815  0.809167748

# Test using aa[5:10, 2] vs aa[5:10, 5]
cov(aa[5:10, 2], aa[5:10, 5])
[1] 0.1644273