Related
I wish to print this data in a table with the columns aligned. I tried with Format but the columns were not aligned. Does anyone know how to do it ? Thank you.
(("tiscali" 10000 2.31 0.84 -14700.0 "none")
("atlantia" 50 22.65 22.68 1.5 "none")
("bper-banca" 1000 1.59 2.01 423.0 "none")
("alerion-cleanpower" 30 44.14 36.45 -230.7 "none")
("tesmec" 10000 0.12 0.14 150.0 "none")
("cover-50" 120 8.95 9.6 78.0 "none")
("ovs" 1000 1.71 1.93 217.0 "none")
("credito-emiliano" 200 5.7 6.26 112.0 "none"))
I tried to align the columns wit the ~T directive, no way. Is there a piece of code that prints nicely table data?
Let's break this down.
First, let's give your data a nice name:
(defparameter *data*
'(("tiscali" 10000 2.31 0.84 -14700.0 "none")
("atlantia" 50 22.65 22.68 1.5 "none")
("bper-banca" 1000 1.59 2.01 423.0 "none")
("alerion-cleanpower" 30 44.14 36.45 -230.7 "none")
("tesmec" 10000 0.12 0.14 150.0 "none")
("cover-50" 120 8.95 9.6 78.0 "none")
("ovs" 1000 1.71 1.93 217.0 "none")
("credito-emiliano" 200 5.7 6.26 112.0 "none")))
Now, come up with a way to print each line using format and destructuring-bind. Widths of various fields are hard-coded in.
(defun print-line (line)
(destructuring-bind (a b c d e f) line
(format T "~20a ~5d ~6,2f ~6,2f ~10,2f ~4a~%" a b c d e f)))
Once you know you can print a line, you just need to do that for each line.
(mapcar 'print-line *data*)
Result:
tiscali 10000 2.31 0.84 -14700.00 none
atlantia 50 22.65 22.68 1.50 none
bper-banca 1000 1.59 2.01 423.00 none
alerion-cleanpower 30 44.14 36.45 -230.70 none
tesmec 10000 0.12 0.14 150.00 none
cover-50 120 8.95 9.60 78.00 none
ovs 1000 1.71 1.93 217.00 none
credito-emiliano 200 5.70 6.26 112.00 none
I have something like this in my personal code, that I reproduced here in a simplified way:
(defpackage :tabular (:use :cl))
(in-package :tabular)
I have a function that turns any object into a list of values (a row), here the usage is for a list of values, so it is already in the correct shape.
(defgeneric columnize (object)
(:documentation "Representation of object as a list of fields")
(:method ((o list)) o))
I also define a transpose method that works with lists of various sizes:
(defun transpose (lists)
(when (notany #'null lists)
(cons
(mapcar #'first lists)
(transpose (mapcar #'cdr lists)))))
Here is your data, as defined by Chris:
(defparameter *data*
'(("tiscali" 10000 2.31 0.84 -14700.0 "none")
("atlantia" 50 22.65 22.68 1.5 "none")
("bper-banca" 1000 1.59 2.01 423.0 "none")
("alerion-cleanpower" 30 44.14 36.45 -230.7 "none")
("tesmec" 10000 0.12 0.14 150.0 "none")
("cover-50" 120 8.95 9.6 78.0 "none")
("ovs" 1000 1.71 1.93 217.0 "none")
("credito-emiliano" 200 5.7 6.26 112.0 "none")))
And finally, a function that prints a list of objects in a tabular way.
Basically, I convert all objects to list of values, convert them to string, and compute their size. This gives a matrix of size that I transpose to have a list of sizes for the same column: this is used to compute the width of each column, based on the maximum size of the actual data.
In practice, I allow also the generic function to add indicators like how to justify (left/right), etc.
(defun tabulate (stream objects)
(loop
for n from 0
for o in objects
for row = (mapcar #'princ-to-string (columnize o))
collect row into rows
collect (mapcar #'length row) into row-widths
finally
(flet ((build-format-arguments (max-width row)
(when (> max-width 0)
(list max-width #\space row))))
(loop
with number-width = (ceiling (log n 10))
with col-widths = (transpose row-widths)
with max-col-widths = (mapcar (lambda (s) (reduce #'max s)) col-widths)
for index from 0
for row in rows
for entries = (mapcan #'build-format-arguments max-col-widths row)
do (format stream
"~v,'0d. ~{~v,,,va~^ ~}~%"
number-width index entries)))))
For example:
(fresh-line)
(tabulate *standard-output* *data*)
Gives:
0. tiscali 10000 2.31 0.84 -14700.0 none
1. atlantia 50 22.65 22.68 1.5 none
2. bper-banca 1000 1.59 2.01 423.0 none
3. alerion-cleanpower 30 44.14 36.45 -230.7 none
4. tesmec 10000 0.12 0.14 150.0 none
5. cover-50 120 8.95 9.6 78.0 none
6. ovs 1000 1.71 1.93 217.0 none
7. credito-emiliano 200 5.7 6.26 112.0 none
As you can see there is some adjustments that could be made to format floating points values so that they align on the dot, but this is already quite useful.
Table 1:
Position
Team
1
MCI
2
LIV
3
MAN
4
CHE
5
LEI
6
AST
7
BOU
8
BRI
9
NEW
10
TOT
Table 2
Position
Team
1
LIV
2
MAN
3
MCI
4
CHE
5
AST
6
LEI
7
BOU
8
TOT
9
BRI
10
NEW
Output I'm looking for is
Position difference = 10 as that is the total of the positional difference. How can I do this in excel/google sheets? So the positional difference is always a positive even if it goes up or down. Think of it as a league table.
Table 2 New (using formula to find positional difference):
Position
Team
Positional Difference
1
LIV
1
2
MAN
1
3
MCI
2
4
CHE
0
5
AST
1
6
LEI
1
7
BOU
0
8
TOT
2
9
BRI
1
10
NEW
1
Try this:
=IFNA(ABS(INDEX(A:B,MATCH(E2,B:B,0),1)-D2),"-")
Assuming that table 1 is at columns A:B:
We currently have a fixed report data that we can only manipulate after download and to simplify, it looks like this:
raw report data extracted to google sheets
a b c
1 Start Date Time Adhering to Schedule (Hours) Time Not Adhering to Schedule (Hours)
2 Employee: A Supervisor: X
3 5/4/2022 7.65 1.35
4 5/5/2022 8.12 0.88
5 5/6/2022 6.95 2.05
6 5/9/2022 8.7 0.3
7 5/10/2022 7.45 1.55
8 5/11/2022 8.63 0.37
9 5/12/2022 8.08 0.92
10 5/13/2022 6.13 0.13
11 Totals: 61.71 7.55
12 Employee: B Supervisor: X
13 5/1/2022 3.8 0.27
14 5/2/2022 6.72 2.28
15 5/3/2022 6.1 2.9
16 5/4/2022 8.43 0.57
17 5/5/2022 5.85 0.53
18 5/10/2022 6.13 2.87
19 5/11/2022 0 1.5
20 5/12/2022 2 1.5
21 5/13/2022 1.75 1.75
22 Totals: 40.78 14.17
I would like some help in constructing a new sheet via formulas so that it rearranges the raw data as follows:
desired output
a b c d e
1 EMPLOYEE SUPERVISOR Start Date Time Adhering to Schedule (Hours) Time Not Adhering to Schedule (Hours)
2 A X 04/05/22 7.65 1.35
3 A X 05/05/22 8.12 0.88
4 A X 06/05/22 6.95 2.05
5 A X 09/05/22 8.70 0.30
6 A X 10/05/22 7.45 1.55
7 A X 11/05/22 8.63 0.37
8 A X 12/05/22 8.08 0.92
9 A X 13/05/22 6.13 0.13
10 B X 01/05/22 3.80 0.27
11 B X 02/05/22 6.72 2.28
12 B X 03/05/22 6.10 2.90
13 B X 04/05/22 8.43 0.57
14 B X 05/05/22 5.85 0.53
15 B X 10/05/22 6.13 2.87
16 B X 11/05/22 0.00 1.50
17 B X 12/05/22 2.00 1.50
18 B X 13/05/22 1.75 1.75
It probably needs some combination of QUERY() ARRAYFORMULA(), TRANSPOSE() and/or INDEX() or something.. but i can't quite figure it out. I need some help with to get started in the right track. the dates and data between employees are dynamic so the formula in the desired result needs to adjust to that as well.
thanks!
edit: adding a sample trix for reference :) https://docs.google.com/spreadsheets/d/1m_FCGcnXvnEiMZ8X4K1eEsMljORWV4V1Yq_81vFnx4Y/edit?usp=sharing
Gobal solution
in E1
={ArrayFormula(if(A1:A="Totals:",,{
substitute(lookup(row(A1:A),row(A1:A)/if(ISNUMBER(A1:A),0,1),A1:A),"Employee: ",""),
substitute(lookup(row(A1:A),row(A1:A)/if(ISNUMBER(A1:A),0,1),C1:C),"Supervisor: ","")
})),Arrayformula(if(ISNUMBER(A1:A),{A1:A,B1:B,C1:C},))}
In 3 steps (3 arrayformulas),
try in H1
=arrayformula(if(left(A1:A,6)="Totals",,if(left(A1:A,8)="Employee",{B1:B,D1:D,E1:E,E1:E,E1:E},{E1:E,E1:E,A1:A,B1:B,C1:C})))
then, back in F1 to complete all rows with employee and supervisor
=ArrayFormula({lookup(row(H:H),row(H:H)/if(H:H<>"",1,0),H:H),lookup(row(I:I),row(I:I)/if(I:I<>"",1,0),I:I)})
finally, if you want to reduce the presentation, in M1
=query(F:L,"select F,G,J,K,L where J is not null",0)
I have a small data set of ~200 samples taken over twenty years with two columns of data that sometimes have multiple entries for the period (i.e. age or date). When I go to plot it, even though the data is over 20 years the graph heavily reflects the number of samples in the period and not the period itself. For example during age 23 there may be 2 or 3 samples, 1 for age 24, 20 for age 25, and 10 for age 35.. the number of samples entirely on needs for additional data at the time.. so simply there is no consistency to the sample rate.
How do I get an Max or an Average / Max for a period (age) and ensure there is only one entry per period in the sheet (about one entry per year) without having to create a separate sheet full of separate queries and charting off of that?
What I have tried in Google Sheets (where my data is) is on the x-series chart choosing "aggregate" (which is on the age period) which helps flatten the graph a bit, but doesn't reduce the series.
A read only link to the the spreadsheet is HERE for reference.
Data Looking something like this:
3/27/2013 36.4247 2.5 29.3
4/10/2013 36.4630 1.8 42.8
4/15/2013 36.4767 2.2 33.9
5/2/2013 36.5233 2.2 33.9
5/21/2013 36.5753 1.91 39.9
5/29/2013 36.5973 1.94 39.2
7/29/2013 36.7644 1.98 38.3
10/25/2013 37.0055 1.7 45.6
2/28/2014 37.3507 1.85 50 41.3
6/1/2014 37.6055 1.98 38 38.1
12/1/2014 38.1068 37
6/1/2015 38.6055 2.18 34 33.9
12/11/2015 39.1342 3.03 23 23.1
12/14/2015 39.1425 3.18 22 21.9
12/15/2015 39.1452 3.44 20 20.0
12/17/2015 39.1507 3.61 19 18.9
12/21/2015 39.1616 3.62 19 18.8
12/23/2015 39.1671 3.32 21 20.8
12/25/2015 39.1726 3.08 23 22.7
12/28/2015 39.1808 3.12 22 22.4
12/29/2015 39.1836 2.97 24 23.7
12/30/2015 39.1863 3.57 19 19.1
12/31/2015 39.1890 3.37 20 20.5
1/1/2016 39.1918 3.37 20 20.5
1/3/2016 39.1973 2.65 27 27.0
1/4/2016 39.2000 2.76 26 25.8
try:
=QUERY(SORTN(SORT({YEAR($A$6:$A), B6:B}, 1, 0, 2, 0), 9^9, 2, 1, 1),
"where Col1 <> 1899")
demo spreadsheet
and build a chart from there
I'm doing some dynamic Monte Carlo simulation in Google Sheets, by utilizing the COUNTIF formula for the simulation. Something is not working the way I thought it would, but I cannot put my finger on. I have two columns that I'm comparing, and I need to count the instances where the value in one column is bigger than the value in the other column. If I do this explicitly by propagating the if comparison formula I obtain the correct result. However, if I do it with
=countif( A4:A, ">" & B4:B )
I do not obtain the correct result. My example is at this sheet, the number in cell C4 is the malfunctioning COUNTIF, which equals 2 in the example, and the number in cell E4 is 5, which is the correct count by propagating the comparison in column F and adding the correct comparisons in E4.
p1 p2 n
0.5 0.51 10
Monte Carlo
0.50 0.60 2 5 0
0.90 0.50 1
0.60 0.30 1
0.50 0.60 0
0.40 0.30 1
0.40 0.50 0
0.60 0.70 0
0.60 0.30 1
0.70 0.50 1
0.10 0.30 0
There are two scenarios with countif:
(1) As a non-array formula, =countif( A4:A, ">" & B4:B ) would give you the same result as =countif( A4:A, ">" & B4 ) i.e. it would count only values of A greater than .60, giving the answer 2.
(2) As an array formula, =sum(countif( A4:A, ">" & B4:B )) would give you a separate result for each value of B (2+5+9+2...) giving the answer 56.
If you wanted to use countif, you would need to do something like this:
=ArrayFormula(countif(A4:A-B4:B,">"&0))
try:
=INDEX(SUM(IF(A4:A>B4:B, 1)))