Stata 如何获得Var_a和Var_b的Var_c之和？
我试图找到两个变量的和 如果我有以下数据：Stata 如何获得Var_a和Var_b的Var_c之和？,stata,Stata,我试图找到两个变量的和 如果我有以下数据： Name Commodity Amount_cmdt Alex apple 5 Ben orange 10 Chris apple 25 Alex orange 10 Alex apple 10 Chris orange 10 Ben a
Name Commodity Amount_cmdt
Alex apple 5
Ben orange 10
Chris apple 25
Alex orange 10
Alex apple 10
Chris orange 10
Ben apple 5
Name Commodity Amount_cmdt total_apple total_orange
Alex apple 5 15 10
Ben orange 10 5 10
Chris apple 25 25 20
Alex orange 10 15 10
Alex apple 10 15 10
Chris orange 10 25 20
Ben apple 5 5 10
Chris orange 10 25 20
foreach var of varlist Name {
foreach var of varlist Commodity {
replace total_apple = sum( Amount_cmdt) if Commodity == "apple"
replace total_orange = sum( Amount_cmdt) if Commodity == "orange"
}
}
list
使用您的玩具示例：
clear
input strL(name commodity) amount total_apple total_orange
Alex apple 5 15 10
Ben orange 10 5 10
Chris apple 25 25 20
Alex orange 10 15 10
Alex apple 10 15 10
Chris orange 10 25 20
Ben apple 5 5 10
Chris orange 10 25 20
end
bysort name commodity: egen totals = total(amount)
bysort name (commodity): generate totalapple = totals[1]
bysort name (commodity): generate totalorange = totals[_N]
list name commodity amount total_apple totalapple total_orange totalorange, abbreviate(15)
+------------------------------------------------------------------------------------+
| name commodity amount total_apple totalapple total_orange totalorange |
|------------------------------------------------------------------------------------|
1. | Alex apple 5 15 15 10 10 |
2. | Alex apple 10 15 15 10 10 |
3. | Alex orange 10 15 15 10 10 |
4. | Ben apple 5 5 5 10 10 |
5. | Ben orange 10 5 5 10 10 |
|------------------------------------------------------------------------------------|
6. | Chris apple 25 25 25 20 20 |
7. | Chris orange 10 25 25 20 20 |
8. | Chris orange 10 25 25 20 20 |
+------------------------------------------------------------------------------------+
编辑： 您可以将其概括为两种以上的商品，如下所示：
clear
input strL(name commodity) amount
Alex apple 5
Ben orange 10
Chris apricot 3
Alex apricot 4
Ben apricot 2
Chris apple 25
Alex orange 10
Alex apple 10
Chris orange 10
Ben apple 5
Chris apricot 15
Alex apricot 6
Chris orange 10
end
bysort name commodity: egen totals = total(amount)
egen commodities = group(commodity)
levelsof commodity, local(allcommodities) clean
local i 0
foreach var of local allcommodities {
local ++i
generate `var' = .
bysort name (commodity): replace `var' = totals if commodities == `i'
bysort name (commodity): egen total`var' = min(`var')
drop `var'
}
drop commodities
list name commodity amount total*, abbreviate(15)
+-------------------------------------------------------------------------------+
| name commodity amount totals totalapple totalapricot totalorange |
|-------------------------------------------------------------------------------|
1. | Alex apple 5 15 15 10 10 |
2. | Alex apple 10 15 15 10 10 |
3. | Alex apricot 6 10 15 10 10 |
4. | Alex apricot 4 10 15 10 10 |
5. | Alex orange 10 10 15 10 10 |
|-------------------------------------------------------------------------------|
6. | Ben apple 5 5 5 2 10 |
7. | Ben apricot 2 2 5 2 10 |
8. | Ben orange 10 10 5 2 10 |
9. | Chris apple 25 25 25 18 20 |
10. | Chris apricot 3 18 25 18 20 |
|-------------------------------------------------------------------------------|
11. | Chris apricot 15 18 25 18 20 |
12. | Chris orange 10 20 25 18 20 |
13. | Chris orange 10 20 25 18 20 |
+-------------------------------------------------------------------------------+
使用您的玩具示例：
clear
input strL(name commodity) amount total_apple total_orange
Alex apple 5 15 10
Ben orange 10 5 10
Chris apple 25 25 20
Alex orange 10 15 10
Alex apple 10 15 10
Chris orange 10 25 20
Ben apple 5 5 10
Chris orange 10 25 20
end
bysort name commodity: egen totals = total(amount)
bysort name (commodity): generate totalapple = totals[1]
bysort name (commodity): generate totalorange = totals[_N]
list name commodity amount total_apple totalapple total_orange totalorange, abbreviate(15)
+------------------------------------------------------------------------------------+
| name commodity amount total_apple totalapple total_orange totalorange |
|------------------------------------------------------------------------------------|
1. | Alex apple 5 15 15 10 10 |
2. | Alex apple 10 15 15 10 10 |
3. | Alex orange 10 15 15 10 10 |
4. | Ben apple 5 5 5 10 10 |
5. | Ben orange 10 5 5 10 10 |
|------------------------------------------------------------------------------------|
6. | Chris apple 25 25 25 20 20 |
7. | Chris orange 10 25 25 20 20 |
8. | Chris orange 10 25 25 20 20 |
+------------------------------------------------------------------------------------+
编辑： 您可以将其概括为两种以上的商品，如下所示：
clear
input strL(name commodity) amount
Alex apple 5
Ben orange 10
Chris apricot 3
Alex apricot 4
Ben apricot 2
Chris apple 25
Alex orange 10
Alex apple 10
Chris orange 10
Ben apple 5
Chris apricot 15
Alex apricot 6
Chris orange 10
end
bysort name commodity: egen totals = total(amount)
egen commodities = group(commodity)
levelsof commodity, local(allcommodities) clean
local i 0
foreach var of local allcommodities {
local ++i
generate `var' = .
bysort name (commodity): replace `var' = totals if commodities == `i'
bysort name (commodity): egen total`var' = min(`var')
drop `var'
}
drop commodities
list name commodity amount total*, abbreviate(15)
+-------------------------------------------------------------------------------+
| name commodity amount totals totalapple totalapricot totalorange |
|-------------------------------------------------------------------------------|
1. | Alex apple 5 15 15 10 10 |
2. | Alex apple 10 15 15 10 10 |
3. | Alex apricot 6 10 15 10 10 |
4. | Alex apricot 4 10 15 10 10 |
5. | Alex orange 10 10 15 10 10 |
|-------------------------------------------------------------------------------|
6. | Ben apple 5 5 5 2 10 |
7. | Ben apricot 2 2 5 2 10 |
8. | Ben orange 10 10 5 2 10 |
9. | Chris apple 25 25 25 18 20 |
10. | Chris apricot 3 18 25 18 20 |
|-------------------------------------------------------------------------------|
11. | Chris apricot 15 18 25 18 20 |
12. | Chris orange 10 20 25 18 20 |
13. | Chris orange 10 20 25 18 20 |
+-------------------------------------------------------------------------------+
@Pearly Spencer向您提供了您所要求的，但详细的代码确实表明，这是一个非常扭曲的数据结构——我预测这将非常难以处理 此外，您不需要重复计算，然后删除重复项，因为您可以直接获得简单的结构 请注意，这会破坏原始数据集，因此保留原始数据集始终是一个好主意。此外，我们不能评论您可能有哪些其他变量。 这两种布局中的一种或两种可能同样或更有用
clear
input str6 (name commodity) amount
Alex apple 5
Ben orange 10
Chris apple 25
Alex orange 10
Alex apple 10
Chris orange 10
Ben apple 5
Chris orange 10
end
collapse (sum) amount, by(name commodity)
list, sepby(name)
+---------------------------+
| name commod~y amount |
|---------------------------|
1. | Alex apple 15 |
2. | Alex orange 10 |
|---------------------------|
3. | Ben apple 5 |
4. | Ben orange 10 |
|---------------------------|
5. | Chris apple 25 |
6. | Chris orange 20 |
+---------------------------+
reshape wide amount, i(name) j(commodity) string
list
+-----------------------------+
| name amoun~le amoun~ge |
|-----------------------------|
1. | Alex 15 10 |
2. | Ben 5 10 |
3. | Chris 25 20 |
+-----------------------------+
@Pearly Spencer向您提供了您所要求的，但详细的代码确实表明，这是一个非常扭曲的数据结构——我预测这将非常难以处理 此外，您不需要重复计算，然后删除重复项，因为您可以直接获得简单的结构 请注意，这会破坏原始数据集，因此保留原始数据集始终是一个好主意。此外，我们不能评论您可能有哪些其他变量。 这两种布局中的一种或两种可能同样或更有用
clear
input str6 (name commodity) amount
Alex apple 5
Ben orange 10
Chris apple 25
Alex orange 10
Alex apple 10
Chris orange 10
Ben apple 5
Chris orange 10
end
collapse (sum) amount, by(name commodity)
list, sepby(name)
+---------------------------+
| name commod~y amount |
|---------------------------|
1. | Alex apple 15 |
2. | Alex orange 10 |
|---------------------------|
3. | Ben apple 5 |
4. | Ben orange 10 |
|---------------------------|
5. | Chris apple 25 |
6. | Chris orange 20 |
+---------------------------+
reshape wide amount, i(name) j(commodity) string
list
+-----------------------------+
| name amoun~le amoun~ge |
|-----------------------------|
1. | Alex 15 10 |
2. | Ben 5 10 |
3. | Chris 25 20 |
+-----------------------------+
您可能会混淆
sum（）foreachsum（）foreach折叠崩溃