R與bp功能循環

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我有一個數據集包含3列：country，year和tdvalue。我想通過使用R breakpoint函數按國家創建一個循環來創建一個具有1或0的虛擬變量（sd），如果年份是斷點。但是，當我讓我的代碼工作時，我的sd變量總是等於0，而我知道這是幾年的情況？R與bp功能循環

非常感謝您的幫助！

library(zoo) 
library(sandwich) 
library(strucchange) 
library(segmented) 
library(tree) 

tabo<-read.table("boucle.txt", header=T, sep="\t") 

Fonction.bp<-function(b) 
    bp.inf <- breakpoints(tabo$year ~ tabo$tradevaluein1000usd , tabo = tabo[b,], h = 8) 
    t<-breakdates(confint(bp.inf)) 
    for (i in 1:nrow(t)) { 
    res <- ifelse(tabo$year[b] == t[i,1] , 1, 0) 
    return(res) 
    } 
} 

numero<-1:nrow(tabo) 
tabo$sd<-lapply(tabo$code_o,Fonction.bp)

數據樣本：

code_o -origin -year -tradevaluein1000usd 

ABW Aruba 1988 375.059 
ABW Aruba 1989 3458.656 
ABW Aruba 1990 2924.484 
ABW Aruba 1991 140509.4

等幾個國家

dput（TABO）：

structure(list(code_o = 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, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L), .Label = c("ABW", "AFG", "AGO"), class = "factor"), 
    origin = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
    3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
    3L, 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, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L), .Label = c("Afghanistan", "Angola", 
    "Aruba"), class = "factor"), year = c(1988L, 1989L, 1990L, 
    1991L, 1992L, 1993L, 1994L, 1995L, 1996L, 1997L, 1998L, 1999L, 
    2000L, 2001L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 
    2009L, 2010L, 2011L, 2012L, 1988L, 1989L, 1990L, 1991L, 1992L, 
    1993L, 1994L, 1995L, 1996L, 1997L, 1998L, 1999L, 2000L, 2001L, 
    2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L, 
    2011L, 2012L, 1988L, 1989L, 1990L, 1991L, 1992L, 1993L, 1994L, 
    1995L, 1996L, 1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L, 
    2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L, 2011L, 2012L 
    ), tradevaluein1000usd = c(375.059, 3458.656, 2924.484, 140509.4, 
    326377, 548739.3, 570287.9, 673563.2, 809647.7, 1021996, 
    680243.7, 944974.8, 1950097, 1416807, 1055372, 1276015, 2503752, 
    3908081, 4294362, 4654180, 5523432, 2203173, 272596.5, 4450387, 
    127760.6, 121861.2, 125059.8, 134163.4, 115283.5, 82499.51, 
    68673.89, 97143.18, 104883.2, 124654.5, 155892.9, 167802.9, 
    137721, 153405.3, 99146.39, 103894.9, 190640.9, 209073.9, 
    264083.6, 254765.3, 408123.6, 507407, 1283451, 609946.1, 
    486418.4, 67638.02, 1112926, 3120863, 4082248, 3290223, 3796494, 
    3283747, 3175830, 3614761, 4669298, 4618304, 3501481, 4478671, 
    7878114, 6290144, 7344164, 8563406, 11900000, 20700000, 30200000, 
    39500000, 65700000, 38900000, 50600000, 59400000, 8839811 
    ), sd = list(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
     0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
     0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
     0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
     0, 0, 0, 0, 0, 0)), .Names = c("code_o", "origin", "year", 
"tradevaluein1000usd", "sd"), row.names = c(NA, -75L), class = "data.frame")

來源

2013-07-22 aleeyah

請輸入您的數據，或將我們指向源代碼，以便您的問題很容易重現。 – ricardo

謝謝，請看現在 – aleeyah

任何幫助pleaseee – aleeyah

你有壞的代碼。

如果你期待別人的時候，你必須投入更多的精力在

您的功能不起作用：tabo['AGO',]，tabo['AFG',]和tabo['ABW',]都是空的，因爲有這些名字沒有行。我想你可能想子集數據，使用類似：

tabo[tabo$code_o == 'AGO',] 
tabo[tabo$code_o == 'AFG',] 
tabo[tabo$code_o == 'ABW',]

bp.inf是一樣的，不管我們是否包含代碼tabo = tabo[b,] - 爲你調用的項目的迴歸，從全球環境，而不是傳入數據幀（因爲您要給tabo而不是data）。如果這是令人困惑的忘記它...

底線是，有幾個erros在行中找到斷點。你需要改變類似bp.inf <- breakpoints(year ~ tradevaluein1000usd, data = tabo[tabo$code_o == 'AGO',], h = 8)。

另請注意，您的功能未打開{，因此根本不起作用。

來源

2013-07-22 21:28:34 ricardo

R與bp功能循環

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