如何将chr转换为数字
我是r的新手，我先用这个表打开文件如何将chr转换为数字,r,type-conversion,numeric,chr,R,Type Conversion,Numeric,Chr,我是r的新手，我先用这个表打开文件 df<- read.csv("geral_modelo_inadimplencia_2.csv",sep=";",stringsAsFactors = FALSE) 如何将chr转换为数字 Saldo.Rotativo A.Vista Parcelado Saque 1 dez/16 2.919.869.506,96 6.364.901.107,55 463.677.827,46
df<- read.csv("geral_modelo_inadimplencia_2.csv",sep=";",stringsAsFactors = FALSE)
Saldo.Rotativo A.Vista Parcelado Saque
1 dez/16 2.919.869.506,96 6.364.901.107,55 463.677.827,46 729.185,18
2 dez/17 2.007.351.784,18 6.831.919.805,09 780.093.428,86 2.817.814,72
3 dez/18 1.944.752.969,68 7.458.695.484,62 542.074.530,21 1.885.633,29
4 jan/19 1.991.796.619,57 7.371.837.099,11 540.893.516,33 2.058.371,60
5 fev/19 2.121.650.035,30 6.967.735.692,25 562.211.487,01 2.132.249,61
6 mar/19 2.062.475.653,11 6.900.028.117,67 575.861.976,61 2.100.849,74
7 abr/19 2.089.793.039,24 6.945.593.710,28 583.181.387,89 2.090.951,69
8 mai/19 2.078.700.800,99 7.146.597.703,16 612.184.578,96 2.132.951,04
9 jun/19 2.239.390.093,82 6.851.118.033,80 618.844.690,37 1.764.866,10
10 jul/19 2.121.263.409,38 7.196.087.606,84 629.995.715,52 3.945.650,40
11 ago/19 2.173.187.280,54 7.089.675.942,22 624.808.459,45 6.341.527,95
12 set/19 2.285.571.063,90 7.111.228.186,19 617.840.220,61 6.143.505,16
13 out/19 2.193.401.889,85 7.263.912.266,04 622.821.392,86 7.253.169,67
14 nov/19 2.281.061.211,60 7.240.713.335,11 611.161.428,40 7.484.398,11
15 dez/19 2.212.531.321,45 7.892.016.606,72 597.916.084,63 6.464.980,78
我们可以使用
str\u remove\u alllibrary(dplyr)
library(stringr)
df <- df %>%
mutate_at(-1, ~ as.numeric(str_replace(str_remove_all(.,
'\\.'), ',', '.')))
df
# date Saldo.Rotativo A.Vista Parcelado Saque
#1 dez/16 2919869507 6364901108 463677827 729185.2
#2 dez/17 2007351784 6831919805 780093429 2817814.7
#3 dez/18 1944752970 7458695485 542074530 1885633.3
#4 jan/19 1991796620 7371837099 540893516 2058371.6
#5 fev/19 2121650035 6967735692 562211487 2132249.6
#6 mar/19 2062475653 6900028118 575861977 2100849.7
#7 abr/19 2089793039 6945593710 583181388 2090951.7
#8 mai/19 2078700801 7146597703 612184579 2132951.0
#9 jun/19 2239390094 6851118034 618844690 1764866.1
#10 jul/19 2121263409 7196087607 629995716 3945650.4
#11 ago/19 2173187281 7089675942 624808459 6341528.0
#12 set/19 2285571064 7111228186 617840221 6143505.2
#13 out/19 2193401890 7263912266 622821393 7253169.7
#14 nov/19 2281061212 7240713335 611161428 7484398.1
#15 dez/19 2212531321 7892016607 597916085 6464980.8
str(df)
#'data.frame': 15 obs. of 5 variables:
# $ date : chr "dez/16" "dez/17" "dez/18" "jan/19" ...
# $ Saldo.Rotativo: num 2.92e+09 2.01e+09 1.94e+09 1.99e+09 2.12e+09 ...
# $ A.Vista : num 6.36e+09 6.83e+09 7.46e+09 7.37e+09 6.97e+09 ...
# $ Parcelado : num 4.64e+08 7.80e+08 5.42e+08 5.41e+08 5.62e+08 ...
# $ Saque : num 729185 2817815 1885633 2058372 2132250 ...
df不幸的是，R似乎没有自动检测不同语言环境的基本函数，因此我们需要一对gsub
s
dat[，-1]as.numeric（sub（“，”，“，”，gsub（“\\.”，”，“2.919.869.506,96”））
akrun，这是一种语言环境。。。一些国家使用
表示千位分隔符，使用，
表示十进制。Marilivia Melo de sousa，您是否意识到StackExchange网站有一种礼仪，即如果答案解决了问题，您应该这样做？这样做不仅给回答者提供了一些观点，也为有类似问题的读者提供了一些结尾。虽然你只能接受一个答案，但你可以选择增加你认为有帮助的选票。（如果仍然存在问题，您可能需要编辑问题并提供更多详细信息。）（您之前的两个问题都有可用的答案。）当然。我认为这是对你的一个“提示”（这是一个很好的建议），但我可以把你的名字从上面去掉。
library(dplyr)
library(stringr)
df <- df %>%
mutate_at(-1, ~ as.numeric(str_replace(str_remove_all(.,
'\\.'), ',', '.')))
df
# date Saldo.Rotativo A.Vista Parcelado Saque
#1 dez/16 2919869507 6364901108 463677827 729185.2
#2 dez/17 2007351784 6831919805 780093429 2817814.7
#3 dez/18 1944752970 7458695485 542074530 1885633.3
#4 jan/19 1991796620 7371837099 540893516 2058371.6
#5 fev/19 2121650035 6967735692 562211487 2132249.6
#6 mar/19 2062475653 6900028118 575861977 2100849.7
#7 abr/19 2089793039 6945593710 583181388 2090951.7
#8 mai/19 2078700801 7146597703 612184579 2132951.0
#9 jun/19 2239390094 6851118034 618844690 1764866.1
#10 jul/19 2121263409 7196087607 629995716 3945650.4
#11 ago/19 2173187281 7089675942 624808459 6341528.0
#12 set/19 2285571064 7111228186 617840221 6143505.2
#13 out/19 2193401890 7263912266 622821393 7253169.7
#14 nov/19 2281061212 7240713335 611161428 7484398.1
#15 dez/19 2212531321 7892016607 597916085 6464980.8
str(df)
#'data.frame': 15 obs. of 5 variables:
# $ date : chr "dez/16" "dez/17" "dez/18" "jan/19" ...
# $ Saldo.Rotativo: num 2.92e+09 2.01e+09 1.94e+09 1.99e+09 2.12e+09 ...
# $ A.Vista : num 6.36e+09 6.83e+09 7.46e+09 7.37e+09 6.97e+09 ...
# $ Parcelado : num 4.64e+08 7.80e+08 5.42e+08 5.41e+08 5.62e+08 ...
# $ Saque : num 729185 2817815 1885633 2058372 2132250 ...
df[-1] <- lapply(df[-1], function(x)
sub(",", ".", gsub(".", "", x, fixed = TRUE)))
df <- type.convert(df, as.is = TRUE)
df <- structure(list(date = c("dez/16", "dez/17", "dez/18", "jan/19",
"fev/19", "mar/19", "abr/19", "mai/19", "jun/19", "jul/19", "ago/19",
"set/19", "out/19", "nov/19", "dez/19"), Saldo.Rotativo = c("2.919.869.506,96",
"2.007.351.784,18", "1.944.752.969,68", "1.991.796.619,57", "2.121.650.035,30",
"2.062.475.653,11", "2.089.793.039,24", "2.078.700.800,99", "2.239.390.093,82",
"2.121.263.409,38", "2.173.187.280,54", "2.285.571.063,90", "2.193.401.889,85",
"2.281.061.211,60", "2.212.531.321,45"), A.Vista = c("6.364.901.107,55",
"6.831.919.805,09", "7.458.695.484,62", "7.371.837.099,11", "6.967.735.692,25",
"6.900.028.117,67", "6.945.593.710,28", "7.146.597.703,16", "6.851.118.033,80",
"7.196.087.606,84", "7.089.675.942,22", "7.111.228.186,19", "7.263.912.266,04",
"7.240.713.335,11", "7.892.016.606,72"), Parcelado = c("463.677.827,46",
"780.093.428,86", "542.074.530,21", "540.893.516,33", "562.211.487,01",
"575.861.976,61", "583.181.387,89", "612.184.578,96", "618.844.690,37",
"629.995.715,52", "624.808.459,45", "617.840.220,61", "622.821.392,86",
"611.161.428,40", "597.916.084,63"), Saque = c("729.185,18",
"2.817.814,72", "1.885.633,29", "2.058.371,60", "2.132.249,61",
"2.100.849,74", "2.090.951,69", "2.132.951,04", "1.764.866,10",
"3.945.650,40", "6.341.527,95", "6.143.505,16", "7.253.169,67",
"7.484.398,11", "6.464.980,78")), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
"14", "15"))