Reemplace la lista de caracteres con una secuencia de flotantes basada en el marco de datos de mapeo
Aug 19 2020
Tengo un marco de datos de mapeo y un marco de datos grande donde cada fila representa proteína con su secuencia.
Quiero una forma eficiente de mapear las secuencias a los valores correspondientes al aminoácido en función del marco de datos de mapeo.
Pude iterar en la secuencia y reemplazarla con el siguiente código:
calcStickiness <- function(seq) {
seq_iter <- strsplit(unlist(seq), "")[[1]]
transformed_seq <- c()
for (c in seq_iter) {
transformed_seq <- c(transformed_seq, stickiness_tabel[stickiness_tabel["X"] == c][2])
}
print(transformed_seq)
}
# calling the function
calcStickiness(row["sequence_full"][1])
donde stickiness_tabelesta:
structure(list(X = c("K", "E", "D", "N", "Q", "S", "P", "R",
"T", "H", "A", "G", "M", "V", "L", "I", "F", "C", "Y", "W"),
x = c(-1.25639466063649, -0.928687786101206, -0.700106643211895,
-0.356971499674196, -0.295054350932285, -0.209468209138379,
-0.177787659972006, -0.0892949396458573, 0.0576667944592403,
0.215277407729333, 0.263739398989502, 0.556792734365241,
0.7448899445842, 0.900506232741908, 1.06680680601946, 1.18416532767113,
1.68723510186035, 1.70109173545121, 1.70150269278206, 2.01452547017961
)), class = "data.frame", row.names = c(NA, -20L))
Quería saber si hay una forma más rápida de hacerlo porque el marco de datos de mi secuencia tiene muchas entradas.
Una fila simple del marco de datos es:
structure(list(X = 1L, code = "12as_1", nsub2 = 2L, pdb_error2 = "NO",
QSBIO_err_prob = 3.5, chain_name = "B", sequence_full = "MKTAYIAKQRQISFVKSHFSRQLEERLGLIEVQAPILSRVGDGTQDNLSGAEKAVQVKVKALPDAQFEVVHSLAKWKRQTLGQHDFSAGEGLYTHMKALRPDEDRLSPLHSVYVDQWDWERVMGDGERQFSTLKSTVEAIWAGIKATEAAVSEEFGLAPFLPDQIHFVHSQELLSRYPDLDAKGRERAIAKDLGAVFLVGIGGKLSDGHRHDVRAPDYDDWSTPSELGHAGLNGDILVWNPVLEDAFELSSMGIRVDADTLKHQLALTGDEDRLELEWHQALLRGEMPQTIGGGIGQSRLTMLLLQLPHIGQVQAGVWPAAVRESVPSLL"), row.names = 1L, class = "data.frame")
Donde estoy interesado sequence_full.
Editar
para la siguiente fila:
MKTAYIAKQRQISFVKSHFSRQLEERLGLIEVQAPILSRVGDGTQDNLSGAEKAVQVKVKALPDAQFEVVHSLAKWKRQTLGQHDFSAGEGLYTHMKALRPDEDRLSPLHSVYVDQWDWERVMGDGERQFSTLKSTVEAIWAGIKATEAAVSEEFGLAPFLPDQIHFVHSQELLSRYPDLDAKGRERAIAKDLGAVFLVGIGGKLSDGHRHDVRAPDYDDWSTPSELGHAGLNGDILVWNPVLEDAFELSSMGIRVDADTLKHQLALTGDEDRLELEWHQALLRGEMPQTIGGGIGQSRLTMLLLQLPHIGQVQAGVWPAAVRESVPSLL
Quiero obtener algo como:
[1] " 0.74488994" "-1.25639466" " 0.05766679" " 0.26373940" " 1.70150269" " 1.18416533" " 0.26373940" "-1.25639466" "-0.29505435"
[10] "-0.08929494" "-0.29505435" " 1.18416533" "-0.20946821" " 1.68723510" " 0.90050623" "-1.25639466" "-0.20946821" " 0.21527741"
[19] " 1.68723510" "-0.20946821" "-0.08929494" "-0.29505435" " 1.06680681" "-0.92868779" "-0.92868779" "-0.08929494" " 1.06680681"
[28] " 0.55679273" " 1.06680681" " 1.18416533" "-0.92868779" " 0.90050623" "-0.29505435" " 0.26373940" "-0.17778766" " 1.18416533"
[37] " 1.06680681" "-0.20946821" "-0.08929494" " 0.90050623" " 0.55679273" "-0.70010664" " 0.55679273" " 0.05766679" "-0.29505435"
[46] "-0.70010664" "-0.35697150" " 1.06680681" "-0.20946821" " 0.55679273" " 0.26373940" "-0.92868779" "-1.25639466" " 0.26373940"
[55] " 0.90050623" "-0.29505435" " 0.90050623" "-1.25639466" " 0.90050623" "-1.25639466" " 0.26373940" " 1.06680681" "-0.17778766"
[64] "-0.70010664" " 0.26373940" "-0.29505435" " 1.68723510" "-0.92868779" " 0.90050623" " 0.90050623" " 0.21527741" "-0.20946821"
[73] " 1.06680681" " 0.26373940" "-1.25639466" " 2.01452547" "-1.25639466" "-0.08929494" "-0.29505435" " 0.05766679" " 1.06680681"
[82] " 0.55679273" "-0.29505435" " 0.21527741" "-0.70010664" " 1.68723510" "-0.20946821" " 0.26373940" " 0.55679273" "-0.92868779"
[91] " 0.55679273" " 1.06680681" " 1.70150269" " 0.05766679" " 0.21527741" " 0.74488994" "-1.25639466" " 0.26373940" " 1.06680681"
[100] "-0.08929494" "-0.17778766" "-0.70010664" "-0.92868779" "-0.70010664" "-0.08929494" " 1.06680681" "-0.20946821" "-0.17778766"
[109] " 1.06680681" " 0.21527741" "-0.20946821" " 0.90050623" " 1.70150269" " 0.90050623" "-0.70010664" "-0.29505435" " 2.01452547"
[118] "-0.70010664" " 2.01452547" "-0.92868779" "-0.08929494" " 0.90050623" " 0.74488994" " 0.55679273" "-0.70010664" " 0.55679273"
[127] "-0.92868779" "-0.08929494" "-0.29505435" " 1.68723510" "-0.20946821" " 0.05766679" " 1.06680681" "-1.25639466" "-0.20946821"
[136] " 0.05766679" " 0.90050623" "-0.92868779" " 0.26373940" " 1.18416533" " 2.01452547" " 0.26373940" " 0.55679273" " 1.18416533"
[145] "-1.25639466" " 0.26373940" " 0.05766679" "-0.92868779" " 0.26373940" " 0.26373940" " 0.90050623" "-0.20946821" "-0.92868779"
[154] "-0.92868779" " 1.68723510" " 0.55679273" " 1.06680681" " 0.26373940" "-0.17778766" " 1.68723510" " 1.06680681" "-0.17778766"
[163] "-0.70010664" "-0.29505435" " 1.18416533" " 0.21527741" " 1.68723510" " 0.90050623" " 0.21527741" "-0.20946821" "-0.29505435"
[172] "-0.92868779" " 1.06680681" " 1.06680681" "-0.20946821" "-0.08929494" " 1.70150269" "-0.17778766" "-0.70010664" " 1.06680681"
[181] "-0.70010664" " 0.26373940" "-1.25639466" " 0.55679273" "-0.08929494" "-0.92868779" "-0.08929494" " 0.26373940" " 1.18416533"
[190] " 0.26373940" "-1.25639466" "-0.70010664" " 1.06680681" " 0.55679273" " 0.26373940" " 0.90050623" " 1.68723510" " 1.06680681"
[199] " 0.90050623" " 0.55679273" " 1.18416533" " 0.55679273" " 0.55679273" "-1.25639466" " 1.06680681" "-0.20946821" "-0.70010664"
[208] " 0.55679273" " 0.21527741" "-0.08929494" " 0.21527741" "-0.70010664" " 0.90050623" "-0.08929494" " 0.26373940" "-0.17778766"
[217] "-0.70010664" " 1.70150269" "-0.70010664" "-0.70010664" " 2.01452547" "-0.20946821" " 0.05766679" "-0.17778766" "-0.20946821"
[226] "-0.92868779" " 1.06680681" " 0.55679273" " 0.21527741" " 0.26373940" " 0.55679273" " 1.06680681" "-0.35697150" " 0.55679273"
[235] "-0.70010664" " 1.18416533" " 1.06680681" " 0.90050623" " 2.01452547" "-0.35697150" "-0.17778766" " 0.90050623" " 1.06680681"
[244] "-0.92868779" "-0.70010664" " 0.26373940" " 1.68723510" "-0.92868779" " 1.06680681" "-0.20946821" "-0.20946821" " 0.74488994"
[253] " 0.55679273" " 1.18416533" "-0.08929494" " 0.90050623" "-0.70010664" " 0.26373940" "-0.70010664" " 0.05766679" " 1.06680681"
[262] "-1.25639466" " 0.21527741" "-0.29505435" " 1.06680681" " 0.26373940" " 1.06680681" " 0.05766679" " 0.55679273" "-0.70010664"
[271] "-0.92868779" "-0.70010664" "-0.08929494" " 1.06680681" "-0.92868779" " 1.06680681" "-0.92868779" " 2.01452547" " 0.21527741"
[280] "-0.29505435" " 0.26373940" " 1.06680681" " 1.06680681" "-0.08929494" " 0.55679273" "-0.92868779" " 0.74488994" "-0.17778766"
[289] "-0.29505435" " 0.05766679" " 1.18416533" " 0.55679273" " 0.55679273" " 0.55679273" " 1.18416533" " 0.55679273" "-0.29505435"
[298] "-0.20946821" "-0.08929494" " 1.06680681" " 0.05766679" " 0.74488994" " 1.06680681" " 1.06680681" " 1.06680681" "-0.29505435"
[307] " 1.06680681" "-0.17778766" " 0.21527741" " 1.18416533" " 0.55679273" "-0.29505435" " 0.90050623" "-0.29505435" " 0.26373940"
[316] " 0.55679273" " 0.90050623" " 2.01452547" "-0.17778766" " 0.26373940" " 0.26373940" " 0.90050623" "-0.08929494" "-0.92868779"
[325] "-0.20946821" " 0.90050623" "-0.17778766" "-0.20946821" " 1.06680681" " 1.06680681"
La salida debe exportarse a un archivo.
Respuestas
1 Edo Aug 19 2020 at 17:31
Llamé a los datos de la misma manera que lo hiciste:
stickiness_tabel <- structure(list(X = c("K", "E", "D", "N", "Q", "S", "P", "R",
"T", "H", "A", "G", "M", "V", "L", "I", "F", "C", "Y", "W"),
x = c(-1.25639466063649, -0.928687786101206, -0.700106643211895,
-0.356971499674196, -0.295054350932285, -0.209468209138379,
-0.177787659972006, -0.0892949396458573, 0.0576667944592403,
0.215277407729333, 0.263739398989502, 0.556792734365241,
0.7448899445842, 0.900506232741908, 1.06680680601946, 1.18416532767113,
1.68723510186035, 1.70109173545121, 1.70150269278206, 2.01452547017961
)), class = "data.frame", row.names = c(NA, -20L))
row <- structure(list(X = 1L, code = "12as_1", nsub2 = 2L, pdb_error2 = "NO",
QSBIO_err_prob = 3.5, chain_name = "B", sequence_full = "MKTAYIAKQRQISFVKSHFSRQLEERLGLIEVQAPILSRVGDGTQDNLSGAEKAVQVKVKALPDAQFEVVHSLAKWKRQTLGQHDFSAGEGLYTHMKALRPDEDRLSPLHSVYVDQWDWERVMGDGERQFSTLKSTVEAIWAGIKATEAAVSEEFGLAPFLPDQIHFVHSQELLSRYPDLDAKGRERAIAKDLGAVFLVGIGGKLSDGHRHDVRAPDYDDWSTPSELGHAGLNGDILVWNPVLEDAFELSSMGIRVDADTLKHQLALTGDEDRLELEWHQALLRGEMPQTIGGGIGQSRLTMLLLQLPHIGQVQAGVWPAAVRESVPSLL"), row.names = 1L, class = "data.frame")
Ahora, lo que podrías hacer es esto:
stickiness <- setNames(stickiness_tabel$x, stickiness_tabel$X)
lapply(strsplit(row$sequence_full, split = ""), function(x) stickiness[x])
Devuelve una lista de vectores numéricos. Cada elemento de la lista corresponde a la fila que convirtió, y cada vector es un vector con nombre de niveles de adherencia nombrados por la letra correspondiente.
¿Es este el resultado que esperabas? Porque no me queda claro a partir de tu pregunta.
1 Humpelstielzchen Aug 19 2020 at 18:10
Tal vez una data.tablesolución se adapte a sus necesidades.
Creé un conjunto de datos de muestra de 1000 filas repitiendo la fila que proporcionó.
library(data.table)
df <- row[rep(1, 1000),] #repeat row
df_dt <- setDT(df) # convert to data.table
value <- setNames(stickiness_tabel$x, stickiness_tabel$X)
start <- Sys.time()
df_dt[, sequence_full := lapply(sequence_full, function(x) value[unlist(strsplit(x, split = ""))])]
end <- Sys.time()
end - start
Time difference of 0.03744602 secs
df_dt[1, sequence_full]
[[1]]
M K T A Y I A K Q
0.74488994 -1.25639466 0.05766679 0.26373940 1.70150269 1.18416533 0.26373940 -1.25639466 -0.29505435
R Q I S F V K S H
-0.08929494 -0.29505435 1.18416533 -0.20946821 1.68723510 0.90050623 -1.25639466 -0.20946821 0.21527741
F S R Q L E E R L
1.68723510 -0.20946821 -0.08929494 -0.29505435 1.06680681 -0.92868779 -0.92868779 -0.08929494 1.06680681
G L I E V Q A P I
0.55679273 1.06680681 1.18416533 -0.92868779 0.90050623 -0.29505435 0.26373940 -0.17778766 1.18416533 ...
Está convirtiendo su tabla adhesiva en un vector e indexándola para sequence_fullcada fila.
Para generarlo puedes hacer:
write.csv(stack(unlist(df_dt[1, sequence_full])), file = "~/sequence_output.csv", row.names = F)
Lo que devuelve un csv con una columna con el valor de adherencia y la otra con el elemento de secuencia.