shift below cells to count for R-CodePudding

I am using the code below to produce following result in Python and I want equivalent for this code on R. here N is the column of dataframe data . CN column is calculated from values of column N with a specific pattern and it gives me following result in python.


 --- ---- 
| N | CN |
 --- ---- 
| 0 |  0 |
| 1 |  1 |
| 1 |  1 |
| 2 |  2 |
| 2 |  2 |
| 0 |  3 |
| 0 |  3 |
| 1 |  4 |
| 1 |  4 |
| 1 |  4 |
| 2 |  5 |
| 2 |  5 |
| 3 |  6 |
| 4 |  7 |
| 0 |  8 |
| 1 |  9 |
| 2 | 10 |
 --- ----

a short overview of my code is

data = pd.read_table(filename,skiprows=15,decimal=',', sep='\t',header=None,names=["Date ","Heure ","temps (s) ","X","Z"," LVDT V(mm) " ,"Force normale (N) ","FT","FN(N) ","TS"," NS(kPa) ","V (mm/min)","Vitesse normale (mm/min)","e (kPa)","k (kPa/mm) " ,"N " ,"Nb cycles normal" ,"Cycles " ,"Etat normal" ,"k imposÈ (kPa/mm)"]) 

data.columns = [col.strip() for col in data.columns.tolist()]    
N = data[data.keys()[15]] 
N = np.array(N) 
data["CN"] = (data.N.shift().bfill() != data.N).astype(int).cumsum()

an example of data.head() is here


 ------- ------------- ------------ ----------- ---------- ---------- ------------ ------------------- ----------- ------------- ----------- ------------ ------------ -------------------------- ------------ ------------ ----- ------------------ -------- ------------- ------------------- ---- 
| Index |    Date     |   Heure    | temps (s) |    X     |  Z(mm)   | LVDT V(mm) | Force normale (N) |    FT     |    FN(N)    | FT (kPa)  |  NS(kPa)   | V (mm/min) | Vitesse normale (mm/min) |  e (kPa)   | k (kPa/mm) |  N  | Nb cycles normal | Cycles | Etat normal | k imposÈ (kPa/mm) | CN |
 ------- ------------- ------------ ----------- ---------- ---------- ------------ ------------------- ----------- ------------- ----------- ------------ ------------ -------------------------- ------------ ------------ ----- ------------------ -------- ------------- ------------------- ---- 
|   184 | 01/02/2022  |  12:36:52  |   402.163 | 6.910243 | 1.204797 | 0.001101   |        299.783665 | 31.494351 | 1428.988908 | 11.188704 | 505.825016 |        0.1 |                      2.0 | 512.438828 |  50.918786 | 0.0 |              0.0 |  Sort  |  Monte      |               0.0 |  0 |
|   185 | 01/02/2022  |  12:36:54  |   404.288 | 6.907822 | 1.205647 | 4.9e-05    |        296.072718 | 31.162313 | 1404.195316 | 11.028167 |  494.97955 |        0.1 |                     -2.0 | 500.084986 |  49.685639 | 0.0 |              0.0 |  Sort  |  Descend    |               0.0 |  0 |
|   186 | 01/02/2022  |  12:36:56  |   406.536 | 6.907906 | 1.204194 | -0.000214  |        300.231424 | 31.586401 | 1429.123486 |  11.21895 | 505.750815 |        0.1 |                      2.0 | 512.370164 |  50.914002 | 0.0 |              0.0 |  Sort  |  Monte      |               0.0 |  0 |
|   187 | 01/02/2022  |  12:36:58  |   408.627 | 6.910751 | 1.204293 | -0.000608  |        300.188686 | 31.754064 | 1428.979519 | 11.244542 | 505.624564 |        0.1 |                      2.0 | 512.309254 |  50.906544 | 0.0 |              0.0 |  Sort  |  Monte      |               0.0 |  0 |
|   188 | 01/02/2022  |  12:37:00  |   410.679 | 6.907805 | 1.205854 | -0.000181  |        296.358074 | 31.563389 | 1415.224427 | 11.129375 | 502.464948 |        0.1 |                      2.0 | 510.702313 |  50.742104 | 0.0 |              0.0 |  Sort  |  Monte      |               0.0 |  0 |
 ------- ------------- ------------ ----------- ---------- ---------- ------------ ------------------- ----------- ------------- ----------- ------------ ------------ -------------------------- ------------ ------------ ----- ------------------ -------- ------------- ------------------- ----

CodePudding user response：

A one line cumsum trick solves it.

cumsum(c(0L, diff(df1$N) != 0))
#>  [1]  0  1  1  2  2  3  3  4  4  4  5  5  6  7  8  9 10

all.equal(
  cumsum(c(0L, diff(df1$N) != 0)),
  df1$CN
)
#> [1] TRUE

^{Created on 2022-02-14 by the reprex package (v2.0.1)}

Data

x <- "
 --- ---- 
| N | CN |
 --- ---- 
| 0 |  0 |
| 1 |  1 |
| 1 |  1 |
| 2 |  2 |
| 2 |  2 |
| 0 |  3 |
| 0 |  3 |
| 1 |  4 |
| 1 |  4 |
| 1 |  4 |
| 2 |  5 |
| 2 |  5 |
| 3 |  6 |
| 4 |  7 |
| 0 |  8 |
| 1 |  9 |
| 2 | 10 |
 --- ---- "
df1 <- read.table(textConnection(x), header = TRUE, sep = "|", comment.char = " ")[2:3]

^{Created on 2022-02-14 by the reprex package (v2.0.1)}