月语言在as.date函数中 [英] month language in the as.date function
问题描述
我不知道如何解决这个问题。这是我的一个df：
I don't know how to figure out this problem. That's a piece of my df:
structure(list(ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L), .Label = c("B0", "B1", "B12", "B2", "B21", "B22", "B26",
"B3", "B33", "B4", "B7", "P1", "P21", "P24", "P24 ", "P25", "P27",
"P28", "P29"), class = "factor"), Date = structure(c(9839, 9946,
10045, 10133, 10190, 10302, 10354, 10423, 10528, 10676, 10756,
10841, 10904, 11032, 11129, 11227, 11290, 11390, 11485, 11571,
11645, 11725, 11843, 11928, 12003, 12128, 12221, 12305, 12380,
12499, 12549, 12640, 12716, 12856, 12926, 12996, 13104, 13580,
13671, 13759, 13802), class = "Date"), T = c(9.6, 10.1, 10.4,
9.9, 9.4, 9.8, 10, 9.8, 9.8, 9.9, 10.3, 10.6, 9.9, 10, 10.3,
10.1, 10.3, 10, 10.2, 10.4, 10.1, 10.1, 10.1, 10.5, 10.3, NA,
NA, NA, NA, 10.3, 10.4, 10.9, 10.6, 10.4, 10.7, 10.2, 10, 10.2,
10.6, 10.5, 10.4), ph = c(6.9, 7.08, 6.96, 7, 7, 6.97, 6.92,
7.02, 6.93, 6.91, 6.83, 6.87, 6.8, 6.92, 7.02, 6.94, 6.94, 6.86,
6.9, 6.89, 6.9, 6.97, 6.92, 6.93, 6.91, 6.88, 6.93, 6.78, 6.87,
6.91, 6.82, 6.91, 6.98, 6.99, 6.79, 6.91, 6.61, 6.86, 6.93, 6.88,
6.74), EC = c(2810, 3020, 2170, 2511, 1695, 3100, 2510, 1759,
1101, 3330, 5370, 3300, 3210, 921, 2300, 3380, 3340, 2850, 3430,
3510, 3450, 3400, 3280, 3170, 3210, 3250, 3010, 2970, 3080, 3120,
3100, 3040, 3100, 2940, 3050, 3070, 3040, 2270, 2990, 2830, 3010
), O2 = c(0.1, 0.1, 1.3, 0.2, 0.2, 0.1, NA, 0.2, 0.1, 0.2, 0.1,
NA, NA, NA, 0.1, 0.1, NA, 0.1, 0.1, 0.2, NA, NA, 0.2, 0.1, 0.1,
0, 0, 0.1, 0.4, 0.2, 0.2, 0.3, 0.2, 0.1, 0.1, 0.3, 0.7, 0.2,
0.4, 0.2, 0.2), Cl = c(696, 718, 722, 856, 776, 752, 745, 788,
822, 727, 650, 800, 766, 700, 800, 720, 760, 710, 730, 720, 810,
610, 720, 830, 820, 740, 670, 510, 710, 500, 640, 630, 650, 430,
660, 660, 630, 560, 680, 670, 670), SO4 = c(152, 111, 133, 245,
194, 110, 105, 104, 185, 156, 137, 194, 196, 170, 220, 230, 240,
200, 220, 200, 220, 170, 230, 210, 240, 280, 240, 190, 260, 360,
280, 250, 220, 380, 240, 240, 230, 320, 220, 210, 220), NO2 = c(NA,
NA, NA, NA, NA, NA, NA, 0.0067, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, 0.015, NA, NA, NA, NA, NA, 0.01,
NA, NA, NA, 0.031, NA, NA, NA, NA, NA, NA, NA), NO3 = c(0.15,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, 0.06, NA, 0.02, NA, 0.02, 0.07, 0.2,
0.02, NA, NA, NA, 0.05, 0.08, NA, NA, NA, NA, NA), Fe = c(22,
20, NA, 23, NA, 25, NA, NA, NA, NA, 27, NA, NA, NA, 32, NA, NA,
NA, 33, NA, NA, NA, 33, NA, NA, NA, 29, NA, NA, NA, 9, NA, NA,
NA, 8.3, NA, NA, NA, 17, NA, NA), Mn = c(3.8, 3.8, NA, 4.5, NA,
4.7, NA, NA, NA, NA, 4.9, NA, NA, NA, 5.8, NA, NA, NA, 6, NA,
NA, NA, 6, NA, NA, NA, 5.3, NA, NA, NA, 4.1, NA, NA, NA, 4.2,
NA, NA, NA, 4.9, NA, NA), Month = c("dicembre", "marzo", "luglio",
"settembre", "novembre", "marzo", "maggio", "luglio", "ottobre",
"marzo", "giugno", "settembre", "novembre", "marzo", "giugno",
"settembre", "novembre", "marzo", "giugno", "settembre", "novembre",
"febbraio", "giugno", "agosto", "novembre", "marzo", "giugno",
"settembre", "novembre", "marzo", "maggio", "agosto", "ottobre",
"marzo", "maggio", "agosto", "novembre", "marzo", "giugno", "settembre",
"ottobre")), .Names = c("ID", "Date", "T", "ph", "EC", "O2",
"Cl", "SO4", "NO2", "NO3", "Fe", "Mn", "Month"), row.names = c(NA,
-41L), class = "data.frame")
我转换日期列转换为Date对象使用：
I convert the Date column into a Date object using:
df$Date<-as.Date(df$Date, "%d.%m.%y")
然后我从Date一个创建Month列，键入： / p>
Then I created the Month column from the Date one, typing:
df$Month <- months(as.Date(df$Date))
但是月份名称在意大利语，当我尝试创建另一列有序因素
But the month names are in Italian and when I try to create another column of ordered factor
df$Month_factor<-factor(df$Month levels=month.name, ordered=T)
月份名称不被识别，只显示列NA。所以我的问题是：是否可以改变创建新列时的几个月的语言？否则，R是否可以将月份认定为有序因素，即使月份名称不是英文？
我需要这几个月作为有序因素，因为我必须绘制如这个发布。
month names are not recognized and it just appears a column NA. So my question is: is it possible to change the language of the months when creating the new column? Else, is it possible for R to recognize the months as an ordered factor even if the month names are not in English? I need the months as an ordered factor because I have to plot values like explained in this post.
推荐答案
它适用于我...没有意大利语，你用 df $ Month< - months（as.Date（df $ Date））你可以给我们你的 sessionInfo（）这是我的
It works for me... there is no italian as you overwrite it with df$Month <- months(as.Date(df$Date)) can you please give us your sessionInfo(), here is mine
sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: x86_64-w64-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=French_France.1252 LC_CTYPE=French_France.1252 LC_MONETARY=French_France.1252
[4] LC_NUMERIC=C LC_TIME=C
attached base packages:
[1] stats graphics grDevices datasets utils methods base
other attached packages:
[1] Rcpp_0.10.4 TeachingDemos_2.9 fastmatch_1.0-4 fasttime_1.0-0 data.table_1.8.9 bit64_0.9-2
[7] bit_1.1-10 vimcom_0.9-8
loaded via a namespace (and not attached):
[1] tools_3.0.1
尝试将 LC_TIME 设置为 C 同 Sys.setlocale（LC_TIME，C）;
这篇关于月语言在as.date函数中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IT屋！
