1

我有两个结构：

max_map <-
structure(list(name = structure(1:11, .Label = c("2-Acetylaminofluorene", 
"amsacrine", "aniline", "aspartame", "cyclophosphamide", "doxorubicin", 
"indomethacin", "phenacetin", "quercetin", "raloxifene", "urethane"
), class = "factor"), value = c(0.811811403850414, 0.8670680916324, 
0.794704077953131, 0.652724115286456, 0.946812003911574, 0.94467294086402, 
0.99210186168903, 0.965998352825426, 0.953645104970837, 0.903845608662668, 
0.858610554863266)), .Names = c("name", "value"), row.names = c(NA, 
-11L), class = "data.frame")

maps <-
structure(list(name = c("2-Acetylaminofluorene", "amsacrine", 
"aniline", "aspartame", "cyclophosphamide", "doxorubicin", "indomethacin", 
"phenacetin", "quercetin", "raloxifene", "urethane"), avg_relations_fan = c(0.596381660936706, 
0.627169363301574, 0.52144016932515, 0.335756276148214, 0.710245148396949, 
0.786168090022777, 0.931928694886563, 0.797790600434933, 0.836458734127729, 
0.764397331494529, 0.548648356310039), baseline = c(0.441175818174093, 
0.661376446637227, 0.470246408568704, 0.325159351267395, 0.664171399502648, 
0.75247341151084, 0.894791275258052, 0.79447733086043, 0.791316894314006, 
0.593161248492605, 0.546928771024265), baseline_mesh = c(0.511440934523423, 
0.635334407445469, 0.466187120416127, 0.292197730456067, 0.712015987803737, 
0.774493950979802, 0.936857915628513, 0.776404901563741, 0.786072875131457, 
0.586564923115283, 0.602183350788001), standard = c(0.441269542443449, 
0.656249151603696, 0.451995996997505, 0.331622681220588, 0.680778834932872, 
0.742015626142688, 0.883911615393179, 0.791293422595675, 0.760673562009157, 
0.559234401021581, 0.555385232882166), sum_relations_fan = c(0.593111715736251, 
0.518197244570419, 0.52676186810563, 0.331234383858585, 0.697489423349489, 
0.77249112456473, 0.940506641487552, 0.79946569580319, 0.82893149142568, 
0.749819491774919, 0.624830313758535), total = c(0.593111715736251, 
0.518197244570419, 0.52676186810563, 0.331234383858585, 0.697489423349489, 
0.77249112456473, 0.940506641487552, 0.79946569580319, 0.82893149142568, 
0.749819491774919, 0.624830313758535)), .Names = c("name", "avg_relations_fan", 
"baseline", "baseline_mesh", "standard", "sum_relations_fan", 
"total"), row.names = c(NA, 11L), class = c("cast_df", "data.frame"
), idvars = "name", rdimnames = list(structure(list(name = c("2-Acetylaminofluorene", 
"amsacrine", "aniline", "aspartame", "cyclophosphamide", "doxorubicin", 
"indomethacin", "phenacetin", "quercetin", "raloxifene", "urethane"
)), .Names = "name", row.names = c("2-Acetylaminofluorene", "amsacrine", 
"aniline", "aspartame", "cyclophosphamide", "doxorubicin", "indomethacin", 
"phenacetin", "quercetin", "raloxifene", "urethane"), class = "data.frame"), 
    structure(list(series = c("avg_relations_fan", "baseline", 
    "baseline_mesh", "standard", "sum_relations_fan", "total"
    )), .Names = "series", row.names = c("avg_relations_fan", 
    "baseline", "baseline_mesh", "standard", "sum_relations_fan", 
    "total"), class = "data.frame")))

我想将该函数应用于x/y数据maps帧，其中 x 是当前值，y 是name.

我已经试过了

mapply(function(x,y) {x/y}, t(maps[,!names(maps) %in% c('name')]), arrange(max_map, name)$value)

但这给了我一个没有任何关联名称的大列表。我希望结果与数据框相似maps，只是具有不同的值。

4

2 回答 2

5

我只是在这里猜测，但也许你想做这样的事情：

m <- merge(maps,max_map)
m[,2:7] <- m[,2:7] / m[,8]

没有merge并且没有指定你有多少列：

maps[,-1] <- maps[,-1] / max_map$value

再次，假设两者的顺序相同。

于 2012-07-18T18:57:05.127 回答
4

Joran 的回答绝对是更好的方法，但这可能会帮助您mapply更好地理解。每个参数都是一个列表，两者中较短的一个被回收，在本例中为第二个。

mapply(function(x,y) {x/y}, maps[,!names(maps) %in% c('name')], list(arrange(max_map, name)$value))

      avg_relations_fan  baseline baseline_mesh  standard sum_relations_fan     total
 [1,]         0.7346308 0.5434462     0.6299997 0.5435616         0.7306028 0.7306028
 [2,]         0.7233219 0.7627734     0.7327388 0.7568600         0.5976431 0.5976431
 [3,]         0.6561438 0.5917252     0.5866172 0.5687601         0.6628403 0.6628403
 [4,]         0.5143923 0.4981574     0.4476589 0.5080595         0.5074646 0.5074646
 [5,]         0.7501438 0.7014818     0.7520141 0.7190222         0.7366715 0.7366715
 [6,]         0.8322119 0.7965438     0.8198541 0.7854736         0.8177339 0.8177339
 [7,]         0.9393478 0.9019147     0.9443163 0.8909484         0.9479940 0.9479940
 [8,]         0.8258716 0.8224417     0.8037332 0.8191457         0.8276057 0.8276057
 [9,]         0.8771174 0.8297813     0.8242824 0.7976485         0.8692243 0.8692243
[10,]         0.8457167 0.6562639     0.6489658 0.6187278         0.8295880 0.8295880
[11,]         0.6389956 0.6369928     0.7013463 0.6468418         0.7277226 0.7277226
于 2012-07-18T19:00:01.787 回答