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Subtract values from different row


By : chuck
Date : September 16 2020, 04:00 AM
it fixes the issue Apologies, misread your formula as having C2-C3, not C2-B3. Slight amendment below to correct for that.
code :
library(dplyr)
library(lubridate)

df <- data.frame(
  NO = c("AAA", "AAA", "AAA", "AAA", "AAA", "BBB", "BBB", "BBB", "BBB", "BBB"),
  T_DATE = dmy_hms(c( "10/08/2019 17:02:00",  "10/08/2019 14:45:00", "10/08/2019 03:23:00",  "09/08/2019 17:06:00", "08/08/2019 23:29:00",  "08/08/2019 09:34:00", "07/08/2019 23:05:00", "07/08/2019 12:07:00", "06/08/2019 22:06:00", "06/08/2019 10:07:00")),
  L_DATE = dmy_hms(c( "10/08/2019 20:35:00", "10/08/2019 15:10:00","10/08/2019 10:25:00", "10/08/2019 01:11:00","09/08/2019 10:27:00", "08/08/2019 21:19:00","08/08/2019 06:09:00","07/08/2019 20:25:00", "07/08/2019 08:53:00", "06/08/2019 19:44:00"))
)

df %>% 
  group_by(NO) %>% 
  mutate(DIFF = difftime(L_DATE, lead(T_DATE), units = "hours"))
#> # A tibble: 10 x 4
#> # Groups:   NO [2]
#>    NO    T_DATE              L_DATE              DIFF           
#>    <fct> <dttm>              <dttm>              <drtn>         
#>  1 AAA   2019-08-10 17:02:00 2019-08-10 20:35:00  5.833333 hours
#>  2 AAA   2019-08-10 14:45:00 2019-08-10 15:10:00 11.783333 hours
#>  3 AAA   2019-08-10 03:23:00 2019-08-10 10:25:00 17.316667 hours
#>  4 AAA   2019-08-09 17:06:00 2019-08-10 01:11:00 25.700000 hours
#>  5 AAA   2019-08-08 23:29:00 2019-08-09 10:27:00        NA hours
#>  6 BBB   2019-08-08 09:34:00 2019-08-08 21:19:00 22.233333 hours
#>  7 BBB   2019-08-07 23:05:00 2019-08-08 06:09:00 18.033333 hours
#>  8 BBB   2019-08-07 12:07:00 2019-08-07 20:25:00 22.316667 hours
#>  9 BBB   2019-08-06 22:06:00 2019-08-07 08:53:00 22.766667 hours
#> 10 BBB   2019-08-06 10:07:00 2019-08-06 19:44:00        NA hours


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Subtract row values in Array2 from specific row values in Array1


By : khayrul
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Your array structure looks slightly different with multiple records, the code works out like this in an ugly manner. I'm assuming you're talking about something like this:
code :
$array1 = array(
    0=>array('product_id'=>4, 'stocks'=>20), 
    1=>array('product_id'=>5, 'stocks'=>60));
$array2 = array(
    0=>array('product_id'=>4, 'quantity'=>3)
    1=>array('product_id'=>5, 'quantity'=>30));
foreach($array1 as $key=>$value){
    foreach($array2 as $key2=>$value2) {
        if($value['product_id']==$value2['product_id']){
            $value['stocks'] -= $value2['quantity'];
            //optimization to avoid searching this again.
            unset($array2[$key]);
        }
    }}

Sum delivered values and subtract it from the received values with the same name mysql


By : Jen Sylvester Fonten
Date : March 29 2020, 07:55 AM
I wish this helpful for you Use this Query. Hope it should be working fine.
SQL FIDDLE DEMO
code :
SELECT
    I.item_name,
    I.Stocked,
    S.Dispensed,
    COALESCE (I.Stocked, 0) - COALESCE (S.Dispensed, 0) AS Remaining
FROM
    (
        SELECT
            item_name AS 'item_name',
            SUM(amount) AS 'Stocked'
        FROM
            com_inv
        GROUP BY
            item_name
    ) I
LEFT JOIN (
    SELECT
        item_name,
        order_status,
        SUM(amount) AS Dispensed
    FROM
        sls_ordrs
    WHERE
        order_status = 'received'
    GROUP BY
        item_name
) S ON I.item_name = S.item_name;

MS ACCESS - cannot subtract returned values from a query from values in another table


By : user2380498
Date : March 29 2020, 07:55 AM
Does that help Your comment states linking fields are LongText which is synonymous with Memo data type, therefore the error message clearly identifies cause. Change the field type to ShortText.
However, really should use ID field in Project table as primary key and save that instead of UR_No into Invoice table. Numbers are more efficient keys.

mysql- subtract values between rows where column values correspond


By : Siska
Date : March 29 2020, 07:55 AM
This might help you I have a table like: , You need a left join:
code :
select 
  t.*,
  t.weight - tt.weight diff
from tablename t left join tablename tt
on tt.number = t.number and tt.event = t.event and tt.weight <> t.weight

How can I cumulatively add or subtract values based on another column values using Pandas?


By : user3310753
Date : March 29 2020, 07:55 AM
around this issue The method to look at a _lagged value is named shift, and then we inspect if the values is positive, negative, or zero with an if-else construct.
So, first we'd construct the column sign. The logic can be packed into 1 line.
code :
df['sign'] = (df.v_out - df.v_out.shift()).apply(lambda x: 0.05 if x > 0 else -0.05 if x < 0 else 0)
df['cumul'] = df.sign.cumsum()
   secs     v_out  sign  cumul                                                                                      
0   0.0 -1.179100  0.00   0.00                                                                                          
1  15.0 -1.179100  0.00   0.00                                                                                          
2  18.0 -1.179200 -0.05  -0.05                                                                                          
3  33.0 -1.181800 -0.05  -0.10                                                                                          
4  48.0  0.029461  0.05  -0.05`   
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