Я пишу функцию, которая пытается добавить значения в одну строку data.frame сразу в несколько столбцов:
require(stringr)
addPointsToKeyRow = function(df, keyRowNum, searchStringForPointColNames, pointsVector){
colsWithMatchingSearchResults = str_match(colnames(df), searchStringForPointColNames)
pointColNums = (which(!is.na(colsWithMatchingSearchResults)))
pointsVectorCleaned = pointsVector[!is.na(pointsVector)]
print(is.vector(pointsVectorCleaned)) #Returns TRUE
print(is.data.frame(pointsVectorCleaned)) #Returns FALSE
print(pointsVectorCleaned)
if(length(pointsVectorCleaned) == length(pointColNums)){
newDf = data.frame(df, stringsAsFactors = FALSE)
newDf[keyRowNum, pointColNums] = as.character(pointsVectorCleaned)
#for(i in 1:length(pointColNums)){
# newDf[keyRowNum,pointColNums[i]]=as.character(pointsVectorCleaned[i])
#}
print(newDf[keyRowNum,])
}
}
Когда я применяю функцию к своим данным ( addPointsToKeyRow(finalDf, which(finalDf[,1]=="key"), "points_q", pointVals)
), я получаю следующие предупреждения:
In
[<-.factor
(*tmp*
, iseq, value = "2"): недопустимый уровень фактора, сгенерировано NA
Я искал ошибку на SO и других сайтах, и, кажется, всегда рекомендуется убедиться, что ваш data.frame имеет stringsAsFactors = FALSE
.
Я думаю, моя проблема может заключаться в том, что когда я подмножество data.frame ( newDf[keyRowNum, pointColNums]
), он больше не сохраняется stringsAsFactors = FALSE
.
Независимо от того, проблема в этом или нет, я бы очень приветствовал помощь в решении этой странной проблемы. Спасибо заранее!
Для примера допустим, что df:
df = structure(list(first = structure(c(7L, 9L, 5L, 4L, 10L, 2L, 3L,
6L, 1L, 8L), .Label = c("autumn", "spring", "summer", "winter",
"july", "betty", "november", "echo", "victor", "tango"), class = "factor"),
last = structure(c(6L, 2L, 4L, 5L, 1L, 8L, 3L, 9L, 10L, 7L
), .Label = c("brummett1", "do", "drorbaugh", "galeno", "gerber",
"key", "lyons", "pecsok", "perezfranco", "swatt"), class = "factor"),
question1 = structure(c(1L, 1L, 1L, 4L, 6L, 2L, 5L, 3L, 5L,
5L), .Label = c("0", "0.25", "1:02:01", "1:2 50%", "2-Jan",
"50%"), class = "factor"), points_q1 = structure(c(1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question2 = structure(c(8L, 10L, 6L, 5L, 2L, 3L, 7L, 1L,
4L, 9L), .Label = c(" a | b; A| Aa | Ab; b| ab | bb; the possibility that the offspring will be heterozygous is about 25%. The same goes for the homozygous recessive it is a 1:1:1:1",
"1/4 heterozygous for \xf1a\xee and 0 recessive for \xf1b\xee",
"16-Mar", "2-Jan", "3:1 25%", "4-Jan", "Male=aabb Female=AAbb Heterozygous is going to be 1/2. Homozygous is going to be 1/4.",
"possible offspring genotypes (each with probability of 0.25): AABb AaBb AAbb Aabb. Question is asking about probability of Aabb_ which is 0.25.",
"The square shows Ab Ab_ Bb Bb so 50% or 1/2. ", "Xa Yb (father) crossed with XA Xb (mother) = 1/2 "
), class = "factor"), points_q2 = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question3 = structure(c(4L, 5L, 3L, 5L, 5L, 5L, 7L, 2L, 6L,
1L), .Label = c("Codominance", "coheritance", "incomplete dominance",
"Incomplete dominance", "Incomplete dominance ", "Incomplete dominance. ",
"Independent Assortment"), class = "factor"), points_q3 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question4 = structure(c(3L, 4L, 2L, 3L, 6L, 3L, 7L, 1L, 5L,
4L), .Label = c("", "co-dominance", "Codominance", "Codominance ",
"Codominance. ", "Codominant ", "Independent Assortment? (Wrong)"
), class = "factor"), points_q4 = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question5 = structure(c(2L, 10L, 6L, 4L, 5L, 3L, 8L, 1L,
7L, 9L), .Label = c(" X | Y; X| XX | XY; x| Xx | xY; the percentage will be 25 % or 1/4 the same applies to the son ",
"0 for daughter_ because male can only give non-colorblind X chromosome (because he's not colorblind an only has one X chromosome). 0.25 for both son and colorblind.",
"0.25", "25% for son and 25% for daughter", "25% for the son and 25% for the daughter ",
"4-Jan", "50%", "Father=XY Mother=X2Y Therefore_ by using the punnet square_ I was able to show/understand that the probability of them having a son AND him being colorblind is 1/4.",
"To have a son or daughter is 50/50. To have a colorblind daughter is .25 whereas to have a colorblind son is .75 because it is carried on the X chromosome and the son is much more likely to inherit this because he has less x to work with",
"XcY (father) XC Xc (mother) Daughter is 1/4 son 1/4"), class = "factor"),
points_q5 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L), .Label = "", class = "factor"), question6 = structure(c(3L,
6L, 7L, 8L, 5L, 2L, 10L, 9L, 4L, 1L), .Label = c("Chromatids ",
"Chromosomes (diploids)", "homologous chromosome pairs",
"Homologous chromosome pairs are being separated. ", "Homologous chromosomes ",
"Homologous pairs ", "homologous pairs of chromosomes", "Homologus Chromosomes ",
"sister chromatids ", "Sister Chromatids?"), class = "factor"),
points_q6 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L), .Label = "", class = "factor"), question7 = structure(c(6L,
8L, 5L, 7L, 8L, 2L, 3L, 1L, 9L, 4L), .Label = c("", "Chromatids (haploids)",
"Daughter Chromosomes?", "One cell to 2", "sister chromatids",
"Sister chromatids", "Sister Chromatids", "Sister chromatids ",
"Sister chromatids within daughter cells are separating. "
), class = "factor"), points_q7 = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question8 = structure(c(1L, 4L, 1L, 2L, 4L, 2L, 3L, 6L, 5L,
3L), .Label = c("sister chromatids", "Sister chromatids",
"Sister Chromatids", "Sister chromatids ", "Sister chromatids are held together by the centromeres. In prophase chromosomes become visible. During metaphase chromosomes attach to spindles. During Anaphase the chromosomes are split apart and in telophase the cells start to create cleavage. ",
"sisters chromatides"), class = "factor"), points_q8 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question9 = structure(c(2L, 4L, 1L, 3L, 4L, 3L, 3L, 2L, 5L,
3L), .Label = c("prohase ", "prophase", "Prophase", "Prophase ",
"They condense during prophase before the rest of the phases. "
), class = "factor"), points_q9 = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question10 = structure(c(1L, 3L, 1L, 2L, 3L, 2L, 2L, 1L,
4L, 2L), .Label = c("anaphase", "Anaphase", "Anaphase ",
"During anaphase. "), class = "factor"), points_q10 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question11 = structure(c(3L, 4L, 3L, 4L, 4L, 4L, 4L, 3L,
1L, 2L), .Label = c("During prophase. ", "Telephase ", "telophase",
"Telophase"), class = "factor"), points_q11 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question12 = structure(c(1L, 3L, 1L, 2L, 3L, 2L, 3L, 1L,
4L, 2L), .Label = c("metaphase", "Metaphase", "Metaphase ",
"Metaphase. "), class = "factor"), points_q12 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"),
question13 = structure(c(1L, 4L, 1L, 4L, 2L, 4L, 2L, 5L,
3L, 6L), .Label = c("centromere", "Centromere", "Centromere. ",
"Centromeres", "centromeres ", "Cleavage"), class = "factor"),
points_q13 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L), .Label = "", class = "factor")), .Names = c("first",
"last", "question1", "points_q1", "question2", "points_q2", "question3",
"points_q3", "question4", "points_q4", "question5", "points_q5",
"question6", "points_q6", "question7", "points_q7", "question8",
"points_q8", "question9", "points_q9", "question10", "points_q10",
"question11", "points_q11", "question12", "points_q12", "question13",
"points_q13"), row.names = c(NA, -10L), class = "data.frame")
which(finalDf[,1]=="key")
равно 1.
pointVals
является c(NA, "2", "2", "2", "2", "2", "2", "2", "1", "1", "1", "1", "1", "1")
Для пояснения я бы хотел, чтобы итоговая таблица выглядела примерно так:
First Last question1 points_q1 question2 points_q2 etc.
key key 0 2 "possible_offspring_genotypes..." 1 etc.
Я сократил вашу функцию, исходя из моего понимания, дайте мне знать, дает ли она то, что вы хотите, или если я что-то неправильно понял
addPointsToKeyRow = function(df, keyRowNum, searchString, pointsVector) {
#Find columns which has searchString in it
cols <- grepl(searchString, colnames(df))
#Check if the columns with searchString and length of pointsVector is the same
if (sum(cols) == length(pointsVector)) {
#Assign the value
df[keyRowNum,cols] <- pointsVector
}
#Return the updated dataframe
df
}
#Convert all the variables in the column from factor to character
df[] <- lapply(df, as.character)
#define the values to be replaced
pointVals <- c("2", "2", "2", "2", "2", "2", "2", "1", "1", "1", "1","1", "1")
#Call the function
df <- addPointsToKeyRow(df, 1, "points_q", pointsval)
#Check the dataframe
df
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я говорю два предложения