Question

我有一个庞大的5,000观测数据集，我的数据子集如下：

AandB

1 222 454 213.51  59.15%
444 630 789.46  6.15%
2 374 798 807.69  32.00%
304 738 263.59  19.95%
177 641 617.86  18.07%
857 937 842.27  51.97%
973 127.33  0.03%
86 205 146.62  1.18%

我需要一个变量中的两个变量A和B。

例如，1 222 454 213.51应该在A列中作为1222454213.51，变量B中的对应观察值应该是59.15% 原始数据中有一个双倍空格，用于分隔A中我想要的值和B中我想要的值。

因此，我需要：

A                    B
1222454213.51     59.15%
444630789.46       6.15%
2374798807.69     32.00%
304738263.59      19.95%
177641617.86      18.07%
857937842.27      51.97%
973127.33          0.03%
86205146.62        1.18%

我能够通过以下方式获取变量A：

generate A = reverse(substr(reverse(AandB),strpos(reverse(AandB), "  "), . ))
replace A = subinstr(A, " ", "", .)

但是，我在提取百分比数字时遇到了麻烦。

Answer 1

一种方法是：

split AandB, p("  ")
rename AandB1 A
rename AandB2 B
replace A = subinstr(A, " ", "", .)

list, separator(0)

     +---------------------------------------------------+
     |                    AandB               A        B |
     |---------------------------------------------------|
  1. | 1 222 454 213.51  59.15%   1222454213.51   59.15% |
  2. |    444 630 789.46  6.15%    444630789.46    6.15% |
  3. | 2 374 798 807.69  32.00%   2374798807.69   32.00% |
  4. |   304 738 263.59  19.95%    304738263.59   19.95% |
  5. |   177 641 617.86  18.07%    177641617.86   18.07% |
  6. |   857 937 842.27  51.97%    857937842.27   51.97% |
  7. |        973 127.33  0.03%       973127.33    0.03% |
  8. |     86 205 146.62  1.18%     86205146.62    1.18% |
     +---------------------------------------------------+

Answer 2

另一种改进的方法是先剥掉最后一个“单词”（Stata意义）：

for (ls in l_of_lists){
  for (sublist in ls){
    if (!ncol(sublist) == 3)
    {
      print(ncol(sublist))
      #sublist <- NULL # this does achieve the desired result
    }
  }
}

如果您确实希望clear input str42 AandB "1 222 454 213.51 59.15%" "444 630 789.46 6.15%" "2 374 798 807.69 32.00%" "304 738 263.59 19.95%" "177 641 617.86 18.07%" "857 937 842.27 51.97%" "973 127.33 0.03%" "86 205 146.62 1.18%" end generate B = word(AandB, -1) generate A = trim(subinstr(AandB, B, "", .)) list AandB A B, separator(0) +------------------------------------------------------+ | AandB A B | |------------------------------------------------------| 1. | 1 222 454 213.51 59.15% 1 222 454 213.51 59.15% | 2. | 444 630 789.46 6.15% 444 630 789.46 6.15% | 3. | 2 374 798 807.69 32.00% 2 374 798 807.69 32.00% | 4. | 304 738 263.59 19.95% 304 738 263.59 19.95% | 5. | 177 641 617.86 18.07% 177 641 617.86 18.07% | 6. | 857 937 842.27 51.97% 857 937 842.27 51.97% | 7. | 973 127.33 0.03% 973 127.33 0.03% | 8. | 86 205 146.62 1.18% 86 205 146.62 1.18% | +------------------------------------------------------+被视为指定了一些非常大的数字，那么

是前进的一种方式。测量到12位有效数字意味着您处于天文学（也许前6位数字是好的）或经济学（并且也许前一位数字是可靠的）。

Answer 3

以下对我有用：

clear
input str50 AandB
"1 222 454 213.51  59.15%"
"444 630 789.46  6.15%"
"2 374 798 807.69  32.00%"
"304 738 263.59  19.95%"
"177 641 617.86  18.07%"
"857 937 842.27  51.97%"
"973 127.33  0.03%"
"86 205 146.62  1.18%"
end

generate A = subinstr(substr(AandB, 1, strpos(AandB,"%")-6)," ", "", .)
generate B = subinstr(substr(AandB, strpos(AandB,"%")-6, .)," ", "", .)

list, separator(0)

     +---------------------------------------------------+
     |                    AandB               A        B |
     |---------------------------------------------------|
  1. | 1 222 454 213.51  59.15%   1222454213.51   59.15% |
  2. |    444 630 789.46  6.15%    444630789.46    6.15% |
  3. | 2 374 798 807.69  32.00%   2374798807.69   32.00% |
  4. |   304 738 263.59  19.95%    304738263.59   19.95% |
  5. |   177 641 617.86  18.07%    177641617.86   18.07% |
  6. |   857 937 842.27  51.97%    857937842.27   51.97% |
  7. |        973 127.33  0.03%       973127.33    0.03% |
  8. |     86 205 146.62  1.18%     86205146.62    1.18% |
     +---------------------------------------------------+

编辑：

再三考虑，可以简化为以下内容：

generate A = subinstr(substr(AandB, 1, strpos(AandB,"  "))," ", "", .)
generate B = subinstr(substr(AandB, strpos(AandB,"  "), .)," ", "", .)

拆分变量以获取最后一个字符串作为新变量

3 个答案: