我正在编写一个函数，尝试一次在data.frame的单行中同时添加多个列的值：
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,])
}
}
将函数应用于数据（addPointsToKeyRow(finalDf, which(finalDf[,1]=="key"), "points_q", pointVals)）时，出现以下警告：
在
[<-.factor（*tmp*中，iseq，值=“ 2”）： 无效的因子水平，NA生成
我一直在SO和其他站点上寻找错误，建议始终似乎是确保您的data.frame具有stringsAsFactors = FALSE。
我认为我的问题可能是当我对data.frame（newDf[keyRowNum, pointColNums]）进行子集设置时，它不再保留stringsAsFactors = FALSE。
不管是不是问题，我都非常欢迎您提供一些帮助来解决这个奇怪的问题。提前非常感谢！
为了举例，假设df为：
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")是1。
pointVals是c(NA, "2", "2", "2", "2", "2", "2", "2", "1", "1", "1", "1",
"1", "1")
为澄清起见，我希望决赛桌看起来像这样：
First Last question1 points_q1 question2 points_q2 etc.
key key 0 2 "possible_offspring_genotypes..." 1 etc.
答案 0 :(得分：2)
根据我的理解，我降低了您的功能，让我知道它是否可以满足您的需求，或者我是否误解了某些内容
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