融化成两个可变的柱子

本文介绍了融化成两个可变的柱子的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 [1]TypeI.altidx06 idx07idx08farve1farve2 如果我这样做： dm 我得到这些变量： TypeI.altvariablevalue 其中idx06， idx08，idx08，farve1，farve2在变量中表示。 但是我真正想要的是这样的： TypeI.altvariablevaluevariable2value2 其中farve1和farve2在变量2和值2中表示。 我想要这样做的原因是我喜欢线条颜色为绿色（如果值下降）和红色（如果上升）。 编辑：Shane展示了如何通过融合的两个融化来重塑数据。但是我的战略接缝从一开始就被认为是不适当的 - 一言以蔽之。 ggplot（dm，aes（x =变量，y =值，组） = type，col = variable2，label = Type，size = I.alt））+ geom_line（）+ geom_text（data = subset（dm，variable ==idx08），hjust = - 0.2，size = 2.5）+ theme_bw（）+ scale_x_discrete（expand = c（0,1））+ opts（legend.position =none） 我假设我需要投射熔化的框架 - 但我无法弄清楚。 数据： d 3L，24L，2L，28L，32L，22L，15L，29L，1L，17L，18L，33L，25L， 13L，30L，11L，26L，9L ，12L，4L，5L，27L，16L，19L，10L，31L， 7L），.Label = c（Alvorligere vold，Andre strafferetligesærlove，Andre tyverier Bedrageri，Brandstiftelse，Butikstyverier mv，Dokumentfalsk，Færdselslovovertræd。iøvrigt，Færdselsuheldmed spiritus，Falsk forklaring iøvrigt，Forbr.mod off。 myndighed mv，Freds- ogærekrænkelser，Hæleri，Hærværk，Indbrud i bank，forretn。mv，Indbrud i fritidshuse，garager mv，Indbrud i villaer ，lejligheder mv，Love vedr。forsvaret og Lign。，Love vedr。spil，bev。，næring，Lov om euforiserende stoffer，Mangler vedkøretøj，Røveri ，Simpel vold，Spiritus- ogpromillekørsel，Trusler，Tyv./brugstyv。af andet，Tyv./brugstyv。af cyke l，Tyv./brugstyv。 af indr。 køretøj，Tyv / brugstyv。 af creallert，Tyveri fra bil，bådm.v.，Ulovlig omgang med hittegods，Våbenloven，Vold o.l. mod关闭。 （16137L，9519L，5930L，5502L，4887L，3582L，3101L， 1738L，1660L，1649L，1551L，1412L，1338L） ，1164L，1154L，1057L， 931L，907L，857L，724L，681L，644L，641L，505L，450L，419L， 405L，328L，324L，324L，320L，281L，262L）， idx06 = c（1,1， 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1， 1,1,1,1,1,1,1,1,1,1,1,1），idx07 = c（0.972675591417568， 0.766866371450899,0.931743805516597,0.813047711781889， 0.88728323699422 ，0.96420233463035，0.855743544078362，1.03710247349823， 0.751470588235294，0.90443686006826，0.808403361344538，0.902834008097166， 0.718181818181818，0.904555314533623，1.02717391304348，0.68957345971564， 1.10324483775811，0.93030303030303，0.805309734513274，0.843003412969283， 0.74914089347079，0.824786324786325 ，1.04060913705584,1.09150326797386， 0.977941176470588,0.892405063291139,0.966666666666667， 0.828125,0.69,0.813559322033898， 0.697841726618705，0.88235294117647， 0.62280701754386），idx08 = C（0.986612873647533，0.712685595207085， 0.840579710144927，0.865628042843233，0.93757225433526，0.823346303501945， 0.905609973285841，1.03356890459364，0.689705882352941，0.909556313993174， 0.798319327731092，0.955465587044534 ，0.714545454545455， 0.620390455531453，1.10869565217391，0.815165876777251，0.64306784660767， 0.818181818181818，0.722713864306785，0.627986348122867， 0.59106529209622，0.927350427350427，1.21319796954315，1.20915032679739， 1.33088235294118，0.759493670886076，1.40833333333333，0.734375， 0.896,0.932203389830508,0.60431654676259,0.872549019607843， 0.675438596491228），farve1 = c（绿色，绿色，绿色，绿色，绿色，绿色 绿色红色绿色绿色绿色绿色绿色绿色红色绿色红色绿色绿色绿色绿色绿色红色红色绿色绿色绿色绿色green，green，green，green，green），farve2 = c（red，green，green，red，红，绿，红，绿，绿，红，绿，，红，绿，绿，红， ，绿色，绿色，绿色，绿色，绿色，红色，红色，红色，红色，绿色，红色，绿色，红色，红色，绿色，绿色，红色）），.Names = c（Type，I.alt，idx06 ，idx07，idx08，farve1，farve2），class =data.frame，row.names = c（NA，-33L）） 解决方案这并不回答你关于剧组的问题，但你可以确定子集并做两个融化，合并： dm1 dm2 colnames（dm2） dm 或者，eq不幸的是，做一个融化（就像你现在这样做），然后将熔化的数据框分两次（％c（idx06，idx07，idx08）作为一个，！idx 作为另一个）并合并该输出。 无论哪种方式，您都可以得到您想要的： >头（dm）类型I.alt变量值variable2 value2 1 Alvorligere vold 1154 idx08 1.108696 farve1 red 2 Alvorligere vold 1154 idx08 1.108696 farve2 red 3 Alvorligere vold 1154 idx07 1.027174 farve1 red 4 Alvorligere vold 1154 idx07 1.027174 farve2 red 5 Alvorligere vold 1154 idx06 1.000000 farve1 red 6 Alvorligere vold 1154 idx06 1.000000 farve2 red I have the following variables in a data frame:[1] "Type" "I.alt" "idx06" "idx07" "idx08" "farve1" "farve2"If I do:dm <- melt(d, id=c("Type","I.alt"))I get these variables:"Type" "I.alt" "variable" "value" Where "idx06", "idx07", "idx08", "farve1", "farve2" are represented in "variable".But what I really want is something like this:"Type" "I.alt" "variable" "value" "variable2" "value2"Where "farve1" and "farve2" are represented in variable2 and value2.The reason I want to do this, is that I'd like something where the line color is green if the value is falling and red if rising.EDIT: Shane has shown how to reshape the data via two melts merged. But my strategy seams to be ill conceived from the beginning - WRONG in one word. See my comment to Shane's solution.ggplot(dm, aes(x=variable,y=value,group=Type,col=variable2, label=Type,size=I.alt))+geom_line()+geom_text(data=subset(dm, variable=="idx08"),hjust=-0.2, size=2.5)+theme_bw()+scale_x_discrete(expand=c(0,1))+opts(legend.position="none")I assume I need to cast the molten frame - but I can't figure it out.The data:d <- structure(list(Type = structure(c(8L, 21L, 23L, 20L, 6L, 14L, 3L, 24L, 2L, 28L, 32L, 22L, 15L, 29L, 1L, 17L, 18L, 33L, 25L, 13L, 30L, 11L, 26L, 9L, 12L, 4L, 5L, 27L, 16L, 19L, 10L, 31L, 7L), .Label = c("Alvorligere vold", "Andre strafferetlige særlove", "Andre tyverier", "Bedrageri", "Brandstiftelse", "Butikstyverier m.v.", "Dokumentfalsk", "Færdselslovovertræd. i øvrigt", "Færdselsuheld med spiritus", "Falsk forklaring i øvrigt", "Forbr. mod off. myndighed m.v.", "Freds- og ærekrænkelser", "Hæleri", "Hærværk", "Indbrud i bank, forretn. m.v.", "Indbrud i fritidshuse, garager mv", "Indbrud i villaer, lejligheder mv", "Love vedr. forsvaret og lign.", "Love vedr. spil, bev., næring", "Lov om euforiserende stoffer", "Mangler ved køretøj", "Røveri", "Simpel vold", "Spiritus- og promillekørsel", "Trusler", "Tyv./brugstyv. af andet", "Tyv./brugstyv. af cykel", "Tyv./brugstyv. af indr. køretøj", "Tyv/brugstyv. af knallert", "Tyveri fra bil, båd m.v.", "Ulovlig omgang med hittegods", "Våbenloven", "Vold o.l. mod off. myndighed"), class = "factor"), I.alt = c(16137L, 9519L, 5930L, 5502L, 4887L, 3582L, 3101L, 1738L, 1660L, 1649L, 1551L, 1412L, 1338L, 1164L, 1154L, 1057L, 931L, 907L, 857L, 724L, 681L, 644L, 641L, 505L, 450L, 419L, 405L, 328L, 324L, 324L, 320L, 281L, 262L), idx06 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), idx07 = c(0.972675591417568, 0.766866371450899, 0.931743805516597, 0.813047711781889, 0.88728323699422, 0.96420233463035, 0.855743544078362, 1.03710247349823, 0.751470588235294, 0.90443686006826, 0.808403361344538, 0.902834008097166, 0.718181818181818, 0.904555314533623, 1.02717391304348, 0.68957345971564, 1.10324483775811, 0.93030303030303, 0.805309734513274, 0.843003412969283, 0.74914089347079, 0.824786324786325, 1.04060913705584, 1.09150326797386, 0.977941176470588, 0.892405063291139, 0.966666666666667, 0.828125, 0.696, 0.813559322033898, 0.697841726618705, 0.88235294117647, 0.62280701754386), idx08 = c(0.986612873647533, 0.712685595207085, 0.840579710144927, 0.865628042843233, 0.93757225433526, 0.823346303501945, 0.905609973285841, 1.03356890459364, 0.689705882352941, 0.909556313993174, 0.798319327731092, 0.955465587044534, 0.714545454545455, 0.620390455531453, 1.10869565217391, 0.815165876777251, 0.64306784660767, 0.818181818181818, 0.722713864306785, 0.627986348122867, 0.59106529209622, 0.927350427350427, 1.21319796954315, 1.20915032679739, 1.33088235294118, 0.759493670886076, 1.40833333333333, 0.734375, 0.896, 0.932203389830508, 0.60431654676259, 0.872549019607843, 0.675438596491228), farve1 = c("green", "green", "green", "green", "green", "green", "green", "red", "green", "green", "green", "green", "green", "green", "red", "green", "red", "green", "green", "green", "green", "green", "red", "red", "green", "green", "green", "green", "green", "green", "green", "green", "green"), farve2 = c("red", "green", "green", "red", "red", "green", "red", "green", "green", "red", "green", "red", "green", "green", "red", "red", "green", "green", "green", "green", "green", "red", "red", "red", "red", "green", "red", "green", "red", "red", "green", "green", "red")), .Names = c("Type", "I.alt", "idx06", "idx07", "idx08", "farve1", "farve2"), class = "data.frame", row.names = c(NA, -33L)) 解决方案 This doesn't answer your question about cast, but you could certainly subset and do two melts, followed by a merge:dm1 <- melt(d[,c("Type","I.alt","idx06","idx07","idx08")], id=c("Type","I.alt"))dm2 <- melt(d[,c("Type","I.alt","farve1","farve2")], id=c("Type","I.alt"))colnames(dm2) <- c("Type", "I.alt", "variable2", "value2")dm <- merge(dm1, dm2)Or, equivalently, do one melt (like you're currently doing) then subset the melted dataframe twice (idx <- variable %in% c("idx06","idx07","idx08") as one and !idx as the other) and merge that output. Either way you get what you want:> head(dm) Type I.alt variable value variable2 value21 Alvorligere vold 1154 idx08 1.108696 farve1 red2 Alvorligere vold 1154 idx08 1.108696 farve2 red3 Alvorligere vold 1154 idx07 1.027174 farve1 red4 Alvorligere vold 1154 idx07 1.027174 farve2 red5 Alvorligere vold 1154 idx06 1.000000 farve1 red6 Alvorligere vold 1154 idx06 1.000000 farve2 red 这篇关于融化成两个可变的柱子的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！