i want know, if can check out if column of data frame starts 0 or 1 , goes till number of rows without breaking sequence. below sample data frame.
structure(list(x = 1:22, snr = c(1.0035798429, 11.9438978154, na, 3.2894877794, 4.0170266411, 1.6310522977, 1.6405414787, 1.6625412522, 0.8489116253, 7.5312259672, 7.2832910726, 0.5732577083, na, 0.8149754292, 1.9981020389, 1.2477052103, 0.9960804911, 10.3402683931, 3.6328270728, 2.5540496855, 41.96873985, 6.2035281045), id = c(109l, 110l, 111l, 112l, 113l, 114l, 116l, 117l, 118l, 119l, 120l, 121l, 123l, 124l, 125l, 126l, 127l, 128l, 130l, 131l, 132l, 133l), signalintensity = c(6.8173738339, 11.5459925418, na, 9.7804203445, 9.8719842219, 9.0781857736, 8.2289312163, 8.0435364446, 6.1793458315, 10.5581798932, 10.4745329822, 4.1572943809, na, 6.0451742752, 8.3100219509, 7.4558770659, 7.1464749962, 11.4284386394, 9.6273795753, 9.6807417299, 13.3364944397, 10.4304671876 )), .names = c("x", "snr", "id", "signalintensity"), class = "data.frame", row.names = c(na, -22l)) how can check columns , return index if present.
edited: sequence looking natural sequence. suppose if data frame has 10 rows, column if present should have sequence 1,2,3,4,5,6,7,8,9,10 or may like0,1,2,3,4,5,6,7,8,9. . sequence starts 0 or 1 , goes till number of rows increment of 1 each row.
you loop through columns sapply. create function check whether there nas. if not (!any), difference (diff) between adjacent element, check if element difference 1 (all(diff(x)==1) , (&) first value of column 0 or 1 (x[1] %in% 0:1). if there na, output column 'false'.
f1 <- function(x) { if(!any(is.na(x))) all(diff(x)==1) & x[1] %in% 0:1 else false} which(sapply(df, f1)) #x #1 
Comments
Post a Comment