如何计算嵌套列表中“子级别”的和

Question

我有一个嵌套列表mylist。在mylist的第一级，有4种植被类型。在第二个“水平”中，我对15个土壤水平(称为level1、level2等)进行了水分胁迫测量，直到level15为止。

这是我的数据集：

mylist <- structure(list(needleleaf_evergreen_temperate_tree = structure(list(
    layer1 = 0.074844339399475, layer2 = 0.10364235153792, layer3 = 0.139199261740378, 
    layer4 = 0.16641474105245, layer5 = 0.169465314138987, layer6 = 0.144044051642624, 
    layer7 = 0.105600767636715, layer8 = 0.0652887370818433, 
    layer9 = 0.0264351101290891, layer10 = 0.00481604897579532, 
    layer11 = 0.000247537572307209, layer12 = 1.73860803440901e-06, 
    layer13 = 4.8438137437183e-10, layer14 = 6.6619638078275e-16, 
    layer15 = 2.88818224958405e-25), .Names = c("layer1", "layer2", 
"layer3", "layer4", "layer5", "layer6", "layer7", "layer8", "layer9", 
"layer10", "layer11", "layer12", "layer13", "layer14", "layer15"
)), needleleaf_evergreen_boreal_tree = structure(list(layer1 = 0.074844339399475, 
    layer2 = 0.10364235153792, layer3 = 0.139199261740378, layer4 = 0.16641474105245, 
    layer5 = 0.169465314138987, layer6 = 0.144044051642624, layer7 = 0.105600767636715, 
    layer8 = 0.0652887370818433, layer9 = 0.0264351101290891, 
    layer10 = 0.00481604897579532, layer11 = 0.000247537572307209, 
    layer12 = 1.73860803440901e-06, layer13 = 4.8438137437183e-10, 
    layer14 = 6.6619638078275e-16, layer15 = 2.88818224958405e-25), .Names = c("layer1", 
"layer2", "layer3", "layer4", "layer5", "layer6", "layer7", "layer8", 
"layer9", "layer10", "layer11", "layer12", "layer13", "layer14", 
"layer15")), needleleaf_deciduous_boreal_tree = structure(list(
    layer1 = 0.074844339399475, layer2 = 0.10364235153792, layer3 = 0.139199261740378, 
    layer4 = 0.16641474105245, layer5 = 0.169465314138987, layer6 = 0.144044051642624, 
    layer7 = 0.105600767636715, layer8 = 0.0652887370818433, 
    layer9 = 0.0264351101290891, layer10 = 0.00481604897579532, 
    layer11 = 0.000247537572307209, layer12 = 1.73860803440901e-06, 
    layer13 = 4.8438137437183e-10, layer14 = 6.6619638078275e-16, 
    layer15 = 2.88818224958405e-25), .Names = c("layer1", "layer2", 
"layer3", "layer4", "layer5", "layer6", "layer7", "layer8", "layer9", 
"layer10", "layer11", "layer12", "layer13", "layer14", "layer15"
)), c3_irrigated = structure(list(layer1 = 0.0754105781789665, 
    layer2 = 0.106401701696586, layer3 = 0.146857331360997, layer4 = 0.181972345543274, 
    layer5 = 0.191196461077081, layer6 = 0.158267532821584, layer7 = 0.094842247909502, 
    layer8 = 0.0370323148644247, layer9 = 0.00750914382571778, 
    layer10 = 0.000504776733855199, layer11 = 5.5627446227042e-06, 
    layer12 = 3.24337357358994e-09, layer13 = 1.50763789751944e-14, 
    layer14 = 2.43174707104567e-23, layer15 = 1.55216144591354e-37), .Names = c("layer1", 
"layer2", "layer3", "layer4", "layer5", "layer6", "layer7", "layer8", 
"layer9", "layer10", "layer11", "layer12", "layer13", "layer14", 
"layer15"))), .Names = c("needleleaf_evergreen_temperate_tree", 
"needleleaf_evergreen_boreal_tree", "needleleaf_deciduous_boreal_tree", 
"c3_irrigated"))

我需要做的是，对每一种植被类型，对15种土壤水平进行水分胁迫的总和(即，将清单中每一“水平”的所有“亚水平”之和)。

实现这一目标的最佳途径是什么？

Answer 1

只需调用lapply，并且因为嵌套内容在列表中，所以运行unlist将子列表项转换为sum的数字向量。

newlist <- lapply(mylist, function(x) sum(unlist(x)))