将0作为填充添加到R中带小数点的数字

Question

我有一个带小数点的数字向量.我想在数字前面添加一个0,在小数位前只有一个整数,同时在有一个的值中保留尾随0.例如,5.466应阅读05.466并43.770应阅读43.770.

这是数据

longM<-c(18.846, 18.906, 5.466, 11.19, 6.894, 7.578, 5.13, 53.868, 57.216, 
0.228, 41.892, 57.576, 44.19, 29.292, 16.452, 48.306, 16.224, 
28.566, 32.328, 25.908, 9.318, 7.332, 38.838, 4.812, 27.588, 
20.922, 30.804, 43.872, 7.734, 41.856, 3.222, 47.364, 5.754, 
21.792, 43.494, 50.82, 48.618, 4.524, 5.454, 48.744, 36.504, 
44.328, 38.61, 39.084, 38.52, 39.504, 4.206, 47.928, 15.996, 
42.99, 46.116, 26.496, 44.148, 40.284, 43.296, 34.248, 22.242, 
29.838, 31.38, 27.12, 22.278, 21.36, 20.142, 15.93, 54.726, 12.072, 
13.05, 11.886, 1.242, 54.42, 48.132, 42.954, 4.986, 32.604, 12.018, 
43.77, 26.172, 30.27, 39.738, 47.514, 27.012, 45.654, 35.796, 
44.52, 30.564)

我试过了

longM2<-formatC(longM,flag=0, width=6,drop0trailing=F)

和

library(stringr)
 longM22<-str_pad(longM,width=6,side="both",pad="0")

Answer 1

你可以使用sprintf:

sprintf("%06.3f",longM)

例如

> sprintf("%06.3f",c(1.0, 33.4, 12.345,0.1243,12000))
[1] "01.000"    "33.400"    "12.345"    "00.124"    "12000.000"

格式%06.3f意味着:

• 你想用零填充数字
• 你想用6个字符格式化数字(*)
• 在点之后你想要3位数的精度

(*)注意,如果数字是>= 100,那么在该点之前将显示超过2位数(基本上打破了格式约束).