levelsof 显示小数点后位数过长

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0621.rar (343 Bytes) 本附件包括：

• 0621.dta

2013-6-21 14:12:42 上传

. levelsof pt1
10.60000038146973 11.60000038146973 11.80000019073486 12.19999980926514 14.39999961853027 14.69999980926514 15.19999980926514


希望得到的结果是
10.6 11.6 11.8 12.2 14.4 14.7 15.2

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关键词：Levels Level Leve EVE ELS 小数点

g double p=round(pt1,.1)
levelsof p
di "`r(levels)'"

sungmoo 发表于 2013-6-21 22:04
g double p=round(pt1,.1)
levelsof p
di "`r(levels)'"

好像还是不行

. clear

. set obs 10
obs was 0, now 10

. gen pt1=10*uniform()

. g double p=round(pt1,.1)

. levelsof p
.5 .9 2 2.9 3.7 3.8 4.100000000000001 5 8.1

. di "`r(levels)'"
.5 .9 2 2.9 3.7 3.8 4.100000000000001 5 8.1

*改成字符可以
g  p=string(round(pt1,.1))
levelsof p
di r(levels)

*改成字符可以

改成字符，要注意其中的引号

clear
set se 10000
set obs 10
g  p=string(round(10*uniform(),.1))

*注意以下的区别：
levelsof p,c
token r(levels)
n di `1'

levelsof p
token r(levels)
n di `1'

levelsof p,c
token `r(levels)'
n di `1'

好像还是不行

这个要看初始数据的精度。随机数生成与其他输入还有些区别。

help data types

Precision of numeric storage types

    floats have about 7 digits of accuracy; the magnitude of the number does not matter.  Thus,
    1234567 can be stored perfectly as a float, as can 1234567e+20.  The number 123456789, however,
    would be rounded to 123456792.  In general, this rounding does not matter.

    If you are storing identification numbers, the rounding could matter.  If the identification
    numbers are integers and take 9 digits or less, store them as longs; otherwise, store them as
    doubles.  doubles have 16 digits of accuracy.

    Stata stores numbers in binary, and this has a second effect on numbers less than 1.  1/10 has
    no perfect binary representation just as 1/11 has no perfect decimal representation.  In float,
    .1 is stored as .10000000149011612.  Note that there are 7 digits of accuracy, just as with
    numbers larger than 1.  Stata, however, performs all calculations in double precision.  If you
    were to store 0.1 in a float called x and then ask, say, "list if x==.1", there would be nothing
    in the list.  The .1 that you just typed was converted to double, with 16 digits of accuracy
    (.100000000000000014...), and that number is never equal to 0.1 stored with float accuracy.

    One solution is to type "list if x==float(.1)".  The float() function rounds its argument to
    float accuracy; see [D] functions.  The other alternative would be store your data as double,
    but this is probably a waste of memory.  Few people have data that is accurate to 1 part in 10
    to the 7th.  Among the exceptions are banks, who keep records accurate to the penny on amounts
    of billions of dollars.  If you are dealing with such financial data, store your dollar amounts
    as doubles.  See float().



******
clear all
set ty float 或 double       /*对应不同结果*/
inp x
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
end

levelsof x
n di "`r(levels)'"

***比较保险且省事的方法是，用group()将各值转换（对应）为整数，再利用整数操作（使用levelsof）。

di  前面的 n 是什么的缩写啊

蓝色 发表于 2013-6-22 13:13
di  前面的 n 是什么的缩写啊

noisily

tostring pt1,replace force usedisplayformat
destring pt1,replace
levelsof pt1

这是我的方法

两步后，其pt1的format与sungmoo的一步同


如果变量很多，小数后有两位，有三位。效果就不一样啦！