r - 根据单独的列计算列中的间隔总和

Question

我有一个数据框 df 看起来像这样:

     X1         X2
1     0 0.06174568
2     0 0.06174568
3     0 0.05978832
4     0 0.05978832
5     0 0.06007480
      ...
22    0 0.06082051
23    0 0.06051641
24  128 0.06329613
25    0 0.06099445
26    0 0.06195348
27    0 0.06022723
28    0 0.06041903
29    0 0.06195348
      ...
36    0 0.06195348
37    0 0.06176168
38    0 0.06233710
39  103 0.06195348
40    0 0.06195348
41    0 0.06387155
42    0 0.06291252

我想总结 X2 的值直到 X1 != 0，即第 1-24 行和第 25-39 行。总和应添加到不同的列 X3:

     X1         X2        X3
1     0 0.06174568 0.0000000
2     0 0.06174568 0.0000000
3     0 0.05978832 0.0000000
4     0 0.05978832 0.0000000
5     0 0.06007480 0.0000000
           ...
22    0 0.06082051 0.0000000
23    0 0.06051641 0.0000000
24  128 0.06329613 1.4660679
25    0 0.06099445 0.0000000
26    0 0.06195348 0.0000000
27    0 0.06022723 0.0000000
28    0 0.06041903 0.0000000
29    0 0.06195348 0.0000000
           ...
36    0 0.06195348 0.0000000
37    0 0.06176168 0.0000000
38    0 0.06233710 0.0000000
39  103 0.06195348 0.9876114
40    0 0.06195348 0.0000000
41    0 0.06387155 0.0000000

使用for循环很容易解决:

j <- 1
df$X3 <- numeric(NROW(df))

for (i in 1:NROW(df)) {
  if (df$X1[i] != 0) {
    df$X3[i] <- sum(df$X2[j:i])
    j <- i
  }
}

但我想知道是否有一种方法可以在不使用循环的情况下计算这些总和，可能是通过使用 apply()？

这是示例数据框:

structure(list(X1 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 128, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 103, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), X2 = c(0.0617456756756757, 
0.0617456756756757, 0.0597883157894737, 0.0597883157894737, 0.0600748, 
0.0611470588235294, 0.0611470588235294, 0.0599241176470588, 0.0606791111111111, 
0.0628226666666667, 0.0599241176470588, 0.0619248571428571, 0.0613493684210526, 
0.0613493684210526, 0.0620369230769231, 0.0594520512820513, 0.0620369230769231, 
0.0608205128205128, 0.0615807692307692, 0.0612766666666667, 0.0608205128205128, 
0.0608205128205128, 0.0605164102564103, 0.0632961290322581, 0.0609944516129032, 
0.0619534838709678, 0.0602272258064516, 0.0604190322580645, 0.0619534838709678, 
0.0623370967741936, 0.0617616774193548, 0.061378064516129, 0.0619534838709678, 
0.0619534838709678, 0.061378064516129, 0.0619534838709678, 0.0617616774193548, 
0.0623370967741936, 0.0619534838709678, 0.0619534838709678, 0.0638715483870968, 
0.0629125161290323, 0.0629125161290323, 0.0629125161290323, 0.0636797419354839, 
0.0638715483870968, 0.0632961290322581, 0.0623370967741936, 0.0652141935483871, 
0.0650223870967742, 0.0632961290322581, 0.0608026451612903, 0.0615698709677419, 
0.0632961290322581, 0.0608604848484849, 0.0648218181818182, 0.0614006666666667, 
0.0596390196078431, 0.0604162926829268, 0.0608205128205128, 0.0608205128205128, 
0.0608205128205128, 0.0595836164383562, 0.0589412054794521, 0.0599826829268293, 
0.0598381463414634, 0.0599826829268293, 0.0598381463414634, 0.0599826829268293, 
0.0599826829268293, 0.0598381463414634, 0.0606138, 0.0598728, 
0.0597246, 0.0597246, 0.0594282, 0.0595764, 0.0601692, 0.0598728, 
0.060021, 0.04617, 0.04617, 0.046398, 0.0603174, 0.0601692, 0.0597246, 
0.0603174, 0.0594282, 0.060021, 0.0601692, 0.0602123076923077, 
0.0606684615384615, 0.0597561538461538, 0.0601005263157895, 0.0607249473684211, 
0.0607249473684211, 0.0600988235294118, 0.0602735294117647, 0.0606229411764706, 
0.0602735294117647, 0.0599241176470588, 0.0614964705882353, 0.0606229411764706, 
0.0606229411764706, 0.0602282857142857, 0.0614158857142857, 0.0602282857142857, 
0.0605676, 0.0598889714285714, 0.0602282857142857, 0.0604713333333333, 
0.0608678666666667, 0.0604713333333333, 0.0600748, 0.0602730666666667, 
0.0602730666666667, 0.0601692, 0.0601692, 0.0598728, 0.0601692, 
0.0601692, 0.0598728, 0.064768, 0.064592, 0.064416, 0.06424, 
0.06424, 0.0642292, 0.0642292, 0.0642292, 0.0647038, 0.0643874, 
0.065026, 0.06403475, 0.06443125, 0.064233, 0.06522425, 0.06443125, 
0.06562075, 0.065026, 0.06443125, 0.06443125, 0.0603493333333333, 
0.0603493333333333, 0.0600195555555556, 0.0603493333333333, 0.0601844444444445, 
0.0601844444444445, 0.0646295, 0.06522425, 0.065026, 0.06522425, 
0.06443125, 0.06443125, 0.0699253333333334, 0.0701013333333333, 
0.0701013333333333, 0.0693973333333333, 0.0697493333333333, 0.0704533333333334, 
0.0699253333333334, 0.0699253333333334, 0.0695733333333333, 0.0697493333333333, 
0.0695997142857143, 0.0701429714285714, 0.0694186285714286, 0.0699618857142857, 
0.0603493333333333, 0.0605142222222222, 0.0610088888888889, 0.0605142222222222, 
0.0606791111111111, 0.0605142222222222)), .Names = c("X1", "X2"
), class = "data.frame", row.names = c(NA, 174L))

Answer 1

我们使用 cumsum 创建一个组，然后执行 sum

library(dplyr)
df1 %>%
    group_by(grp = lag(cumsum(X1 !=0), default = 0)) %>% 
    mutate(X3 = replace(rep(sum(X2),n()),  X1==0, 0))