我正在尝试从data.frame dat中仅提取32种特定物种，并将所有data.frame的另一个species创建到一个col中，而我还提取year，values和temperature，并将它们放在单个列中。我还将放置每个月的月份。
data.frame的示例：
structure(list(Year = c(1994L, 1995L, 1996L, 1997L, 1998L, 1999L,
2000L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L,
2010L, 2011L, 2012L, 2013L), Species = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), .Label = "Blackbird", class = "factor"), Farmland = c(96.0309523809524,
96.8520833333333, 96.781746031746, 96.8597222222222, 97.4410299003322,
96.6654846335697, 96.858803986711, 97.0811403508772, 96.9259974259974,
97.2803571428571, 96.6017598343685, 96.3777777777778, 96.3227670288895,
96.8100546279118, 96.431746031746, 96.6232323232323, 96.2537878787879,
96.1431827431827, 96.0778288740245), X.Jan. = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L), .Label = "Jan", class = "factor"), atwo.TempJanuary = c(5.06916107894286,
4.390669300225, 3.88357903166667, 1.80642228995455, 5.16489863448837,
5.54367179174468, 4.83031500397674, 5.40830211455263, 4.26790743608108,
4.927588606725, 5.841963431, 4.3303368412, 7.08188921457143,
6.75067792993878, 2.83417096753488, 1.36880495640909, 4.35569636247727,
5.82305364068889, 3.52697043756522)), row.names = c(NA, -19L), class = "data.frame")
一个额外的示例（这是原始的data.frame dat）：
structure(list(Year = c(2006L, 2007L, 1999L, 2004L, 1995L, 2011L,
2011L), Species = structure(c(2L, 4L, 3L, 6L, 2L, 5L, 1L), .Label = c("Buzzard",
"Collared Dove", "Greenfinch", "Linnet", "Meadow Pipit", "Willow Warbler"
), class = "factor"), TempJanuary = c(2.128387049, 4.233225712,
5.270967624, 4.826451505, 4.390322483, 3.841290237, 3.981290234
), TempFebruary = c(0.927499979, 3.098928502, 4.67428561, 5.05103437,
6.343214144, 6.414285571, 6.625356995), TempMarch = c(1.637741899,
3.22096767, 7.312257901, 6.444515985, 5.337096655, 6.787741784,
7.052903068), TempApril = c(4.877333224, 5.888999868, 9.510666454,
9.386333124, 9.005333132, 12.40966639, 12.50166639), TempMay = c(8.729999805,
7.748064343, 13.09096745, 12.1638707, 11.68935458, 12.83032229,
13.07967713), TempJune = c(11.48033308, 11.20633308, 13.91166636,
15.77399965, 14.05266635, 14.30733301, 14.56133301), TempJuly = c(14.86354805,
11.9338707, 17.85612863, 16.44451576, 18.92935442, 15.53612868,
15.75161255), TempAugust = c(12.45225779, 11.48419329, 16.54935447,
18.31516088, 19.22483828, 15.80225771, 16.08387061), TempSeptember = c(13.45633303,
10.09333311, 15.94333298, 15.27299966, 13.52733303, 15.41933299,
15.68566632), TempOctober = c(10.24387074, 7.462903059, 10.5161288,
10.84709653, 13.05967713, 12.67774165, 12.83967713), TempNovember = c(4.650999896,
3.614999919, 7.246333171, 7.388666502, 7.455999833, 9.371333124,
9.511333121), TempDecember = c(3.764516045, 2.116774146, 4.268064421,
4.825161182, 2.01741931, 5.582903101, 5.701290195), Farmland = c(100L,
100L, 40L, 90L, 80L, 10L, 80L)), row.names = c(1L, 100L, 1000L,
2000L, 3000L, 5000L, 10000L), class = "data.frame")
再看看data.frame：
'data.frame': 19 obs. of 5 variables:
$ Year : int 1994 1995 1996 1997 1998 1999 2000 2002 2003 2004 ...
$ Species : Factor w/ 1 level "Blackbird": 1 1 1 1 1 1 1 1 1 1 ...
$ Farmland : num 96 96.9 96.8 96.9 97.4 ...
$ X.Jan. : Factor w/ 1 level "Jan": 1 1 1 1 1 1 1 1 1 1 ...
$ atwo.TempJanuary: num 5.07 4.39 3.88 1.81 5.16 ...
深入了解dat：
Year Species TempJanuary TempFebruary TempMarch TempApril
1 2006 Collared Dove 2.128387 0.927500 1.637742 4.877333
100 2007 Linnet 4.233226 3.098929 3.220968 5.889000
1000 1999 Greenfinch 5.270968 4.674286 7.312258 9.510666
2000 2004 Willow Warbler 4.826452 5.051034 6.444516 9.386333
3000 1995 Collared Dove 4.390322 6.343214 5.337097 9.005333
5000 2011 Meadow Pipit 3.841290 6.414286 6.787742 12.409666
10000 2011 Buzzard 3.981290 6.625357 7.052903 12.501666
TempMay TempJune TempJuly TempAugust TempSeptember TempOctober
1 8.730000 11.48033 14.86355 12.45226 13.45633 10.243871
100 7.748064 11.20633 11.93387 11.48419 10.09333 7.462903
1000 13.090967 13.91167 17.85613 16.54935 15.94333 10.516129
2000 12.163871 15.77400 16.44452 18.31516 15.27300 10.847097
3000 11.689355 14.05267 18.92935 19.22484 13.52733 13.059677
5000 12.830322 14.30733 15.53613 15.80226 15.41933 12.677742
10000 13.079677 14.56133 15.75161 16.08387 15.68567 12.839677
TempNovember TempDecember Farmland
1 4.651000 3.764516 100
100 3.615000 2.116774 100
1000 7.246333 4.268064 40
2000 7.388667 4.825161 90
3000 7.456000 2.017419 80
5000 9.371333 5.582903 10
10000 9.511333 5.701290 80
这是我一直在这里使用的一些代码示例：
#Blackbird population-------------------------------------------------------------
Black_Bird<-aggregate(Farmland ~ Year + Species + TempJanuary, dat[dat$Species=="Blackbird" & dat$Farmland >80,],mean)
Black_bird <- ddply(Black_Bird, .(Year, Species, TempJanuary), Farmland=round(mean(Farmland), 2))
aone<-aggregate(Farmland ~ Year + Species, Black_bird, mean)
atwo<-aggregate(TempJanuary ~ Year + Species, Black_bird, mean)
aone<-aone[, -2]
#Buzzard Population-----------
Buzzard_Bird <-aggregate(Farmland ~ Year + Species + TempJanuary, dat[dat$Species=="Buzzard" & dat$Farmland >80,],mean)
Buzzard_bird <- ddply(Buzzard_Bird, .(Year, Species, TempJanuary), Farmland=round(mean(Farmland), 2))
athree<-aggregate(Farmland ~ Year + Species, Buzzard_bird, mean)
afour<-aggregate(TempJanuary ~ Year + Species, Buzzard_bird, mean)
athree<-athree[, -2]
#Combine and melt into single columns-----------------------------------------------------
mod1<-cbind(atwo, afour, aone, athree)
melt(mod1, id.vars = c("Year", "Farmland", "Species"), measure.vars = c("TempJanuary"), variable.name = "Month", value.name = "Temperature" )
melt的工作效率不高，似乎没有将秃鹰与Blackbird放在同一列中。它停在19行并切断。这似乎无效且耗时。有没有更快，更有效的解决方案？
它应该是这样的：
Year Species Farmland Month Temperature
2008 Blackbird 83.0 Jan 9.011174
2009 Blackbird 83.0 Jan 10.155201
2012 Greenfinch 83.0 Feb 9.578269
2009 Swallow 83.0 Mar 10.361573
2010 Robin 84.5 Oct 9.191641
我有32种可供选择：
[1] Dunnock Blackbird Song Thrush Bullfinch
[5] Corn Bunting Turtle Dove Grey Partridge Yellow Wagtail
[9] Starling Linnet Yellowhammer Skylark
[13] Kestrel Reed Bunting Whitethroat Greenfinch
[17] Rook Stock dove Goldfinch Woodpigeon
[21] Jackdaw House martin Swallow Lapwing
[25] Wren Robin Blue Tit Great tit
[29] Long-tailed Tit Chaffinch Buzzard Sparrowhawk
32 Levels: Blackbird Blue Tit Bullfinch Buzzard ... Yellowhammer
以及从1月到12月的12个月的气温。
以下是一些先前的代码，这些代码使我误入歧途：
library(psych)
dat_two <- aggregate(Farmland ~ Species + Year + TempJanuary + TempFebruary + TempMarch + TempApril + TempMay + TempJune + TempJuly + TempAugust + TempSeptember + TempOctober + TempNovember + TempDecember, dat[dat$Species %in% c('Starling', 'Skylark', 'Yellow Wagtail', 'Kestrel', 'Yellowhammer', 'Greenfinch', 'Swallow', 'Lapwing', 'House Martin', 'Long-tailed Tit', 'Linnet', 'Grey Partridge', 'Turtle Dove', 'Corn Bunting', 'Bullfinch', 'Song Thrush', 'Blackbird', 'Dunnock', 'Whitethroat', 'Rook', 'Woodpigeon', 'Reed Bunting', 'Stock Dove', 'Goldfinch', 'Jackdaw', 'Wren', 'Robin', 'Blue Tit', 'Great Tit', 'Chaffinch', 'Buzzard', 'Sparrowhawk') & dat$Farmland >80,], mean)
dat_three <- aggregate(Farmland ~ Species + Year + TempJanuary + TempFebruary + TempMarch + TempApril + TempMay + TempJune + TempJuly + TempAugust + TempSeptember + TempOctober + TempNovember + TempDecember , dat_two, mean)
colnames(dat_two) <- c("Species", "Year", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec", "Farmland")
library(plyr)
dat_one <- ddply(dat_three, .(Species, Year, TempJanuary, TempFebruary, TempMarch, TempApril, TempMay, TempJune, TempJuly, TempAugust, TempSeptember, TempOctober, TempNovember, TempDecember), summarise, mean = round(mean(Farmland), 2))
#-----------------------------------------------------------------
Jan_Year <- ddply(dat_one, .(Year), summarise, TempJanuary=round(geometric.mean(TempJanuary, na.rm=TRUE), 2))
Feb_Year <- ddply(dat_one, .(Year), summarise, TempFebruary=round(geometric.mean(TempFebruary, na.rm=TRUE), 2))
Mar_Year <- ddply(dat_one, .(Year), summarise, TempMarch=round(geometric.mean(TempMarch, na.rm=TRUE), 2))
Apr_Year <- ddply(dat_one, .(Year), summarise, TempApril=round(geometric.mean(TempApril, na.rm=TRUE), 2))
May_Year <- ddply(dat_one, .(Year), summarise, TempMay=round(geometric.mean(TempMay, na.rm=TRUE), 2))
Jun_Year <- ddply(dat_one, .(Year), summarise, TempJune=round(geometric.mean(TempJune, na.rm=TRUE), 2))
Jun_Year <- ddply(dat_one, .(Year), summarise, TempJune=round(geometric.mean(TempJune, na.rm=TRUE), 2))
Jul_Year <- ddply(dat_one, .(Year), summarise, TempJuly=round(geometric.mean(TempJuly, na.rm=TRUE), 2))
Aug_Year <- ddply(dat_one, .(Year), summarise, TempAugust=round(geometric.mean(TempAugust, na.rm=TRUE), 2))
Sep_Year <- ddply(dat_one, .(Year), summarise, TempSeptember=round(geometric.mean(TempSeptember, na.rm=TRUE), 2))
Oct_Year <- ddply(dat_one, .(Year), summarise, TempOctober=round(geometric.mean(TempOctober, na.rm=TRUE), 2))
Nov_Year <- ddply(dat_one, .(Year), summarise, TempNovember=round(geometric.mean(TempNovember, na.rm=TRUE), 2))
Dec_Year <- ddply(dat_one, .(Year), summarise, TempDecember=round(geometric.mean(TempDecember, na.rm=TRUE), 2))
Farm_Year <- ddply(dat_one, .(Year), summarise, Farmland=round(geometric.mean(mean, na.rm=TRUE), 2))
Farm_Temp <- cbind(Farm_Year, Jan_Year, Feb_Year, Mar_Year, Apr_Year,May_Year, Jun_Year, Jul_Year, Aug_Year, Sep_Year, Oct_Year, Nov_Year, Dec_Year)
Farm_Temp <- Farm_Temp[, !duplicated(colnames(Farm_Temp))]
colnames(Farm_Temp) <- c("Year", "Farmland", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec")
Farm_Temp <- Farm_Temp[, -2]
#-----------------------------
Spring <- aggregate((TempMarch + TempApril + TempMay)/3~Year, Farm_Temp, mean)
Summer <- aggregate((TempJune + TempJuly + TempAugust)/3 ~ Year, Farm_Temp, geometric.mean)
Autumn <- aggregate((TempSeptember + TempOctober+TempNovember)/3~Year, Farm_Temp, geometric.mean)
Winter <- aggregate((TempDecember + TempJanuary + TempFebruary)/3~Year, Farm_Temp, geometric.mean)
Season_Temp <- cbind(Farm_Year, Spring,Summer, Autumn, Winter)
Season_Temp <- Season_Temp[, !duplicated(colnames(Season_Temp))]
colnames(Season_Temp) <- c("Year", "Farmland", "spring", "Summer", "Autumn", "Winter")
#--------------------------------------------------------------------------------------------------------------
library(reshape2)
Season_practice <- aggregate((Mar+ Apr + May)/3 ~ Year + Species + Farmland, dat_two, geometric.mean)
prac1 <- aggregate((Jun+ Jul + Aug)/3 ~ Year + Species + Farmland, dat_two, geometric.mean)
prac1 <- prac1[, c(-1, -2, -3)]
prac2 <- aggregate((Sep + Oct + Nov)/3 ~ Year + Species + Farmland, dat_two, geometric.mean)
prac2 <- prac2[, c(-1, -2, -3)]
prac3 <- aggregate((Dec+ Jan + Feb)/3 ~ Year + Species + Farmland, dat_two, geometric.mean)
prac3 <- prac3[, c(-1, -2, -3)]
Season_practice <- cbind(Season_practice, prac1, prac2, prac3)
colnames(Season_practice) <- c("Year", "Species", "Farmland", "Spring", "Summer", "Autumn", "Winter")
Seasonal_Temp <- melt(Season_practice, id.vars = c("Year", "Species", "Farmland"), measure = c("Spring", "Summer", "Autumn", "Winter"), variable.name = "Month", value.name = "Temperature")
Practicing_Temp <- melt(dat_two, id.vars = c("Year", "Species"), measure = c('Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec'), variable.name = "Month", value.name = "Temperature")
这是从较大的data.frame中摘录的内容，旨在完成所提及的内容。正如您所看到的，与价值有关的季节在重复，因为月份具有不同的价值，所以情况并非如此，所以我在路上一定会出错：
Year Species Month Farmland
1 1994 Blackbird Spring 95.96875
2 1995 Blackbird Spring 95.46875
3 1996 Blackbird Spring 95.64815
4 1997 Blackbird Spring 95.62071
5 1998 Blackbird Spring 95.71925
6 1999 Blackbird Spring 95.74444
7 2000 Blackbird Spring 95.82440
8 2002 Blackbird Spring 95.78333
9 2003 Blackbird Spring 95.61640
10 2004 Blackbird Spring 95.86797
11 2005 Blackbird Spring 95.08452
12 2006 Blackbird Spring 94.66667
13 2007 Blackbird Spring 95.60745
14 2008 Blackbird Spring 93.98383
15 2009 Blackbird Spring 95.08167
16 2010 Blackbird Spring 95.23426
17 2011 Blackbird Spring 95.25000
18 2012 Blackbird Spring 94.75204
19 2013 Blackbird Spring 94.28821
20 1994 Blackbird Summer 95.96875
21 1995 Blackbird Summer 95.46875
22 1996 Blackbird Summer 95.64815
23 1997 Blackbird Summer 95.62071
24 1998 Blackbird Summer 95.71925
25 1999 Blackbird Summer 95.74444
26 2000 Blackbird Summer 95.82440
27 2002 Blackbird Summer 95.78333
28 2003 Blackbird Summer 95.61640
29 2004 Blackbird Summer 95.86797
30 2005 Blackbird Summer 95.08452
31 2006 Blackbird Summer 94.66667
32 2007 Blackbird Summer 95.60745
33 2008 Blackbird Summer 93.98383
34 2009 Blackbird Summer 95.08167
35 2010 Blackbird Summer 95.23426
36 2011 Blackbird Summer 95.25000
37 2012 Blackbird Summer 94.75204
38 2013 Blackbird Summer 94.28821
39 1994 Blackbird Autumn 95.96875
40 1995 Blackbird Autumn 95.46875
41 1996 Blackbird Autumn 95.64815
42 1997 Blackbird Autumn 95.62071
43 1998 Blackbird Autumn 95.71925
44 1999 Blackbird Autumn 95.74444
45 2000 Blackbird Autumn 95.82440
46 2002 Blackbird Autumn 95.78333
47 2003 Blackbird Autumn 95.61640
48 2004 Blackbird Autumn 95.86797
49 2005 Blackbird Autumn 95.08452
50 2006 Blackbird Autumn 94.66667
51 2007 Blackbird Autumn 95.60745
52 2008 Blackbird Autumn 93.98383
53 2009 Blackbird Autumn 95.08167
54 2010 Blackbird Autumn 95.23426
55 2011 Blackbird Autumn 95.25000
56 2012 Blackbird Autumn 94.75204
57 2013 Blackbird Autumn 94.28821
58 1994 Blackbird Winter 95.96875
59 1995 Blackbird Winter 95.46875
60 1996 Blackbird Winter 95.64815
61 1997 Blackbird Winter 95.62071
62 1998 Blackbird Winter 95.71925
63 1999 Blackbird Winter 95.74444
64 2000 Blackbird Winter 95.82440
65 2002 Blackbird Winter 95.78333
66 2003 Blackbird Winter 95.61640
67 2004 Blackbird Winter 95.86797
68 2005 Blackbird Winter 95.08452
69 2006 Blackbird Winter 94.66667
70 2007 Blackbird Winter 95.60745
71 2008 Blackbird Winter 93.98383
72 2009 Blackbird Winter 95.08167
73 2010 Blackbird Winter 95.23426
74 2011 Blackbird Winter 95.25000
75 2012 Blackbird Winter 94.75204
76 2013 Blackbird Winter 94.28821
答案 0 :(得分：2)
考虑reshape将数据的格式从宽格式转换为长格式，然后按年份，月份或指定的季节aggregate重新构建。
输入
Year,Species,TempJanuary,TempFebruary,TempMarch,TempApril,TempMay,TempJune,TempJuly,TempAugust,TempSeptember,TempOctober,TempNovember,TempDecember,Farmland
2006,Collared Dove,2.128387,0.9275,1.637742,4.877333,8.73,11.48033,14.86355,12.45226,13.45633,10.243871,4.651,3.764516,100
2007,Linnet,4.233226,3.098929,3.220968,5.889,7.748064,11.20633,11.93387,11.48419,10.09333,7.462903,3.615,2.116774,100
1999,Greenfinch,5.270968,4.674286,7.312258,9.510666,13.090967,13.91167,17.85613,16.54935,15.94333,10.516129,7.246333,4.268064,40
2004,Willow Warbler,4.826452,5.051034,6.444516,9.386333,12.163871,15.774,16.44452,18.31516,15.273,10.847097,7.388667,4.825161,90
1995,Collared Dove,4.390322,6.343214,5.337097,9.005333,11.689355,14.05267,18.92935,19.22484,13.52733,13.059677,7.456,2.017419,80
2011,Meadow Pipit,3.84129,6.414286,6.787742,12.409666,12.830322,14.30733,15.53613,15.80226,15.41933,12.677742,9.371333,5.582903,10
2011,Buzzard,3.98129,6.625357,7.052903,12.501666,13.079677,14.56133,15.75161,16.08387,15.68567,12.839677,9.511333,5.70129,80
R
bird_df = read.csv(...)
# RESHAPE WIDE TO LONG
r_df <- reshape(bird_df, varying = colnames(bird_df)[3:14], times = colnames(bird_df)[3:14],
v.names = "Temperature", timevar = "Month",
new.row.names = 1:1E5, direction = "long")
# ASSIGN COLUMNS
r_df$Month <- factor(substr(gsub("Temp", "", r_df$Month), 1, 3), levels = month.abb)
r_df$Season <- ifelse(r_df$Month %in% c("Mar", "Apr", "May"), "Spring",
ifelse(r_df$Month %in% c("Jun", "Jul", "Aug"), "Summer",
ifelse(r_df$Month %in% c("Sep", "Oct", "Nov"), "Autumn",
ifelse(r_df$Month %in% c("Dec", "Jan", "Feb"), "Winter", NA)
)
)
)
# RE-ORDER ROWS
r_df <- data.frame(with(r_df, r_df[order(Year, Month, Species),]),
row.names = NULL)
输出
head(r_df)
# Year Species Farmland Month Temperature id Season
# 1 1995 Collared Dove 80 Jan 4.390322 5 Winter
# 2 1995 Collared Dove 80 Feb 6.343214 5 Winter
# 3 1995 Collared Dove 80 Mar 5.337097 5 Spring
# 4 1995 Collared Dove 80 Apr 9.005333 5 Spring
# 5 1995 Collared Dove 80 May 11.689355 5 Spring
# 6 1995 Collared Dove 80 Jun 14.052670 5 Summer
# ...
aggregate(cbind(Temperature, Farmland) ~ Species + Year, r_df, mean)
# Year Species Temperature Farmland
# 1 2011 Buzzard 11.114639 80
# 2 1995 Collared Dove 10.419384 80
# 3 2006 Collared Dove 7.434402 100
# 4 1999 Greenfinch 10.512513 40
# 5 2007 Linnet 6.841882 100
# ...
aggregate(cbind(Temperature, Farmland) ~ Species + Year + Month, r_df, mean)
# Year Month Species Temperature Farmland
# 1 2011 Jan Buzzard 3.981290 80
# 2 2011 Feb Buzzard 6.625357 80
# 3 2011 Mar Buzzard 7.052903 80
# 4 2011 Apr Buzzard 12.501666 80
# 5 2011 May Buzzard 13.079677 80
# ...
aggregate(cbind(Temperature, Farmland) ~ Species + Year + Season, r_df, mean)
# Species Year Season Temperature Farmland
# 1 Collared Dove 1995 Autumn 11.347669 80
# 2 Greenfinch 1999 Autumn 11.235264 40
# 3 Willow Warbler 2004 Autumn 11.169588 90
# 4 Collared Dove 2006 Autumn 9.450400 100
# 5 Linnet 2007 Autumn 7.057078 100
# ...
答案 1 :(得分：1)
我认为这就是您要的？您必须安装tidyverse
library('tidyverse')
dat %>%
pivot_longer(matches('Temp'),
names_to = 'Month',
values_to = 'Temp',
names_prefix = 'Temp')