R:dplyr 按組匯總數據，使用 nth() 調用，在聚合期間計算變量 n

Question

我正在聚合具有可變 bin 大小的數據（請參閱此處的上一個問題： R：根據第二列的 sum(n) 將每 n 行與變量 n 聚合）。 除了計算變量范圍組的總和和均值之外，我還需要在每個組范圍的中點提取單值協變量。 當我嘗試即時執行此操作時，我只能獲得第一組的值和其余組的 NA。

df.summary<-as.data.frame(df %>% 
  mutate(rn = row_number()) %>%
  group_by(grp = (cumsum(d)-1)%/% 100 + 1) %>%
  summarise(x=mean(x, na.rm = TRUE), d=sum(d, na.rm=T), ,i.start=first(rn), i.end=last(rn), y=nth(y, round(first(rn)+(last(rn)-first(rn))/2-1))))

head(df.summary)
  grp          x         d i.start i.end        y
1   1 0.07458317  88.99342       1     4 19.78992
2   2 0.07594546  97.62130       5     8       NA
3   3 0.05353308 104.69683       9    12       NA
4   4 0.06498291 106.23468      13    16       NA
5   5 0.08601759  98.24939      17    20       NA
6   6 0.06262427  84.43745      21    23       NA

樣本數據：

structure(list(x = c(0.10000112377193, 0.110742170350877, 0.0300274304561404, 
0.0575619395964912, 0.109060465438596, 0.0595491225614035, 0.0539270264912281, 
0.0812452063859649, 0.0341699389122807, 0.0391744879122807, 0.0411787485614035, 
0.0996091644385965, 0.0970479474912281, 0.0595715843684211, 0.0483489989122807, 
0.0549631194561404, 0.0705080555964912, 0.080437472631579, 0.105883664631579, 
0.0872411613684211, 0.103236660631579, 0.0381296894912281, 0.0465064491578947, 
0.0936565184561403, 0.0410095752631579, 0.0311180032105263, 0.0257758157894737, 
0.0354721928947368, 0.0584999394736842, 0.0241286060175439, 0.112053376666667, 
0.0769823868596491, 0.0558137530526316, 0.0374491000701754, 0.0419279142631579, 
0.0260257506842105, 0.0544360374561404, 0.107411071842105, 0.103873468, 
0.0419322114035088, 0.0483912961052632, 0.0328373653157895, 0.0866868717719298, 
0.063990467245614, 0.0799280314035088, 0.123490407070175, 0.145676836280702, 
0.0292878782807018, 0.0432093036666667, 0.0203547443684211), 
    d = c(22.2483512600033, 22.2483529247042, 22.2483545865809, 
    22.2483562542823, 22.24835791863, 25.1243105415557, 25.1243148759953, 
    25.1243192107884, 25.1243235416981, 25.1243278750792, 27.2240858553058, 
    27.2240943134697, 27.2241027638674, 27.224111222031, 27.2241196741942, 
    24.5623431981188, 24.5623453409221, 24.5623474809012, 24.562349626705, 
    24.5623517696847, 28.1458125837154, 28.1458157376341, 28.1458188889053, 
    28.1458220452951, 28.1458251983314, 27.8293318542146, 27.8293366652115, 
    27.8293414829159, 27.829346292148, 27.8293511094993, 27.5271773325046, 
    27.5271834011289, 27.5271894694002, 27.5271955369655, 27.5272016048837, 
    28.0376097925214, 28.0376146410729, 28.0376194959786, 28.0376243427651, 
    28.0376291969647, 26.8766095768196, 26.8766122563318, 26.8766149309023, 
    26.8766176123562, 26.8766202925746, 27.8736950101666, 27.8736960528853, 
    27.8736971017815, 27.8736981446767, 27.8736991932199), y = c(19.79001, 
    19.789922, 19.789834, 19.789746, 19.789658, 19.78957, 19.789468, 
    19.789366, 19.789264, 19.789162, 19.78906, 19.78896, 19.78886, 
    19.78876, 19.78866, 19.78856, 19.788458, 19.788356, 19.788254, 
    19.788152, 19.78805, 19.787948, 19.787846, 19.787744, 19.787642, 
    19.78754, 19.787442, 19.787344, 19.787246, 19.787148, 19.78705, 
    19.786956, 19.786862, 19.786768, 19.786674, 19.78658, 19.786486, 
    19.786392, 19.786298, 19.786204, 19.78611, 19.786016, 19.785922, 
    19.785828, 19.785734, 19.78564, 19.785544, 19.785448, 19.785352, 
    19.785256)), row.names = c(NA, 50L), class = "data.frame")

Answer 1

df %>% 
 mutate(rn = row_number()) %>%
 group_by(grp = (cumsum(d)-1)%/% 100 + 1) %>%
 summarise(x=mean(x, na.rm = TRUE),
           d = sum(d, na.rm=T),
           i.start=first(rn), 
           i.end=last(rn), 
           y = nth(y, floor(median(rn)) - i.start))

Answer 2

讓我們在summarise部分添加變量z和n 。 這些變量定義如下。

df %>% 
  mutate(rn = row_number()) %>%
  group_by(grp = (cumsum(d)-1)%/% 100 + 1) %>%
  summarise(x=mean(x, na.rm = TRUE), 
            d=sum(d, na.rm=T), ,i.start=first(rn), 
            i.end=last(rn), 
            z = round(first(rn)+(last(rn)-first(rn))/2-1),
            n = n())

     grp      x     d i.start i.end     z     n
   <dbl>  <dbl> <dbl>   <int> <int> <dbl> <int>
 1     1 0.0746  89.0       1     4     2     4
 2     2 0.0759  97.6       5     8     6     4
 3     3 0.0535 105.        9    12    10     4
 4     4 0.0650 106.       13    16    14     4
 5     5 0.0860  98.2      17    20    18     4
 6     6 0.0626  84.4      21    23    21     3
 7     7 0.0479 112.       24    27    24     4
 8     8 0.0394  83.5      28    30    28     3
 9     9 0.0706 110.       31    34    32     4
10    10 0.0575 112.       35    38    36     4
11    11 0.0647  83.0      39    41    39     3
12    12 0.0659 108.       42    45    42     4
13    13 0.0854 111.       46    49    46     4
14    14 0.0204  27.9      50    50    49     1

在上面的數據框中， n表示由grp分隔的每個組的樣本大小。 但是，當您聲明group_by(grp) ，當您調用nth(y, z) ，您將調用 Z-TH VALUE BY GROUP 。

這意味着對於第 5 組，雖然只有 4 個值，但您調用y第 18 個值。 所以它打印NA 。

為了簡單起見，我認為最簡單的方法是使用n() 。

df %>% 
  mutate(rn = row_number()) %>%
  group_by(grp = (cumsum(d)-1)%/% 100 + 1) %>%
  summarise(x=mean(x, na.rm = TRUE), 
            d=sum(d, na.rm=T), ,i.start=first(rn), 
            i.end=last(rn), 
            y=nth(y, round(n()/2)))

     grp      x     d i.start i.end     y
   <dbl>  <dbl> <dbl>   <int> <int> <dbl>
 1     1 0.0746  89.0       1     4  19.8
 2     2 0.0759  97.6       5     8  19.8
 3     3 0.0535 105.        9    12  19.8
 4     4 0.0650 106.       13    16  19.8
 5     5 0.0860  98.2      17    20  19.8
 6     6 0.0626  84.4      21    23  19.8
 7     7 0.0479 112.       24    27  19.8
 8     8 0.0394  83.5      28    30  19.8
 9     9 0.0706 110.       31    34  19.8
10    10 0.0575 112.       35    38  19.8
11    11 0.0647  83.0      39    41  19.8
12    12 0.0659 108.       42    45  19.8
13    13 0.0854 111.       46    49  19.8
14    14 0.0204  27.9      50    50  NA  

您將調用floor(n/2) th y ，這意味着y位於每個組的中間。 請注意，您也可以嘗試floor(n/2)+1 。