sum | A budgeting app built on Sinatra | Command Line Interface library
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kandi X-RAY | sum Summary
A budgeting app built on Sinatra (view it at sumapp.com)
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QUESTION
I have the data I'm querying
...ANSWER
Answered 2021-Jun-16 at 01:56You cannot reference a column alias in the SELECT
where it is defined. The reason is simple: SQL does not guarantee the order of evaluation of expressions in the SELECT
.
You can use a CTE, subquery, or repeat the expression:
QUESTION
I need to retrieve a range delimited by indexes from a specific array
I cannot use OFFSET because it doesnt use an array as a parameter. And the range will then be use for a secondary calculation
The example:
I want to calculate the SUM of the 4th to the 11th value in the column Numbers.
So at the end the formula should look something like:
=SUM(Numbers[4:10]) = 36
4 and 10 being the desired indexes.
I tried with OFFSET and INDEX but cant figure out how to do it.
...ANSWER
Answered 2021-Jun-16 at 02:36Index()
, INDIRECT()
will work. Also OFFSET()
will work but need to apply some trick. As @BigBen suggested you can use INDEX()
like =SUM(INDEX(B:B,5):INDEX(B:B,11))
but I am sure you will not prefer to hard code index no. So, you can use below formula to dynamically input two index and get sum between those index. Try-
QUESTION
var Employees = [
{
"id": "382740",
"PayrollID": "8117817425",
"EmployeeName": "Bob Jones",
"StartTime": "15:15:00.0000000",
"FinishTime": "18:15:00.0000000",
"BreakTime": "45",
"TotalTime": 2,
"Comments": "Test",
"Rate": "19"
},
{
"id": "439617",
"PayrollID": "8117817425",
"EmployeeName": "Peter Pan",
"StartTime": "16:15:00.0000000",
"FinishTime": "21:15:00.0000000",
"BreakTime": "60",
"TotalTime": 4,
"Comments": "Test",
"Rate": "32"
},
{
"id": "201636",
"PayrollID": "5042289623",
"EmployeeName": "Bob Jones",
"StartTime": "09:56:00.0000000",
"FinishTime": "11:56:00.0000000",
"BreakTime": "45",
"TotalTime": 1.25,
"Comments": "Test Comments",
"Rate": "19"
},
{
"id": "799653",
"PayrollID": "5042289623",
"EmployeeName": "Clarke Kent",
"StartTime": "16:49:00.0000000",
"FinishTime": "21:49:00.0000000",
"BreakTime": "60",
"TotalTime": 4,
"Comments": "Test",
"Rate": "19"
},
{
"id": "951567",
"PayrollID": "5042289623",
"EmployeeName": "Bob Jones",
"StartTime": "01:49:00.0000000",
"FinishTime": "16:49:00.0000000",
"BreakTime": "60",
"TotalTime": 14,
"Comments": "Test",
"Rate": "10"
}
]
...ANSWER
Answered 2021-Jun-16 at 02:44In the Map, set the value not to the cumulative total time for the employee so far, but to a whole employee object that contains the total time inside it. Spread the first object found so as not to mutate the input.
QUESTION
I have a data frame including three columns named 'Altitude', 'Distance', 'Slope'. The column of 'Slope' is calculated using the two first columns 'Altitude', 'Distance'. @ the first step the purpose was to calculate 'Slope' using a condition explained below: A condition function was deployed to start from the top column of the "Distance" variable and add up (sum) values until the summation of them is greater or equal to 10 (>=10). If this condition corrects then calculate the "Slope" using the given formula: Slope=Average(Altitude)/(sum(Distance)). The summation of the 'Distance' was counting from the first value of that to the index that the 'Distance' has stopped there). The following code is for the above explanation (By Tim Roberts):
...ANSWER
Answered 2021-May-19 at 13:38Use this code after you calculate s
to get slope column with desired values:
QUESTION
I have two tables as follows:
...ANSWER
Answered 2021-Jun-15 at 19:02select user_id,name
, count(case when col_a = true then 1 end)
+ count(case when col_b = true then 1 end) total
from tableA a
join TableB b on a.user_id= b.id
group by user_id,name
QUESTION
Hi I tired to check null values of my data frame(house) which has 81 columns but
house.isnull().sum()
display only few columns data.
ANSWER
Answered 2021-Jun-15 at 23:08Try running this line before you get the output
pandas.set_option('display.max_rows', 500)
See this other article on this
QUESTION
I see that jq
can calculate addition as simply as jq 'map(.duration) | add'
but I've got a more complex command and I can't figure out how to perform this add
at the end of it.
I'm starting with data like this:
...ANSWER
Answered 2021-Jun-15 at 22:54If any of your output is going to be raw, you need to pass -r
; it'll just be ignored for data items that aren't strings.
Anyhow -- if you write (expr1, expr2)
, then your input will be passed through both expressions. Thus:
QUESTION
Is it possible to two reduce objects into one by summing their properties like so for any generic object
...ANSWER
Answered 2021-Jun-04 at 20:04A functional approach would be (but probably not clean)
QUESTION
entry = [["D 300"],["D 300"],["W 200"],["D 100"]]
def bankbalance(entry):
deposits = [float(entry[ent][0][2:]) for ent in entry if ("D" in entry[ent][0])]
withdrawals = [float(entry[ent][0][2:]) for ent in entry if ("W" in entry[ent][0])]
global balance
balance = sum(deposits) - sum(withdrawals)
bankbalance(entry)
Print(f'Current balance is {balance}')
...ANSWER
Answered 2021-Jun-15 at 11:02ent
is not the index, it is an element of entry, so you don't need entry[ent][0][2:]
, what you need is ent[0][2:]
.
Fixed code:
QUESTION
Details
I'm working on an algo dealing with a multi-dimensional array. If there is a zero, then the elements of the same column, but following arrays will also equal zero. I want to be able to sum the items that are not zeroed out.
ANSWER
Answered 2021-Jun-15 at 17:18Try this code
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