PYTHON
I am making a calculator that displays formulas and answers of a few different things after you enter time, distance, mass, etc. Velocity works fine, but Speed and Acceleration are display apostrophes, commas, and parentheses. Some of them aren't even in the editor.
Output:
('Speed = Distance / Time = 1m / 1s', ' = 1m/s')
('Acceleration = Force / Mass = 1N', ' / 1kg = 1.0m/s/s')
Velocity = Speed + Direction = 1m/s + Direction = 1m/s North
The variables in the program:
SpeedFormula = 'Speed = Distance / Time = ' + distanceDisplay + ' / ' + timeDisplay, ' = ' + speedDisplay
AccelerationFormula = 'Acceleration = Force / Mass = ' + forceDisplay, ' / ' + massDisplay + ' = ' + AccelerationDisplay
VelocityFormula = 'Velocity = Speed + Direction = ' + speedDisplay + " + " + 'Direction' + ' = ' + VelocityDisplay
Anyone know why they display differently and how I can fix it?
Replace
timeDisplay, ' = ' + speedDisplay
with
timeDisplay + ' = ' + speedDisplay
When you use comma SpeedFormula becomes a tuple, not a string.
P.S. You should probably use formatting like this:
distanceDisplay = 20
timeDisplay = 2
speedDisplay = 10
SpeedFormula = 'Speed = Distance / Time = %d / %d = %d' % (distanceDisplay,timeDisplay,speedDisplay)
Related
Perplexed by the function of using a second = sign in vba. eg. s = Int(xVal) + (xVal = n + 1)
I had been deciphering some code and came across the following line which is perplexing me somewhat and despite some extensive research and debugging I seem to be struggling to find the answer:
s = Int(xVal) + (xVal = n + 1)
and
p(i, 3) = A(i)(s + 3 + (s = n)) + (s = n) * (p(i, 1) - p(i, 2))
My question is what is the function of the comparisons within the parentheses after the first assignment = sign?
TIA
(s = n)
If both s and n have the same value then this evaluates to True, which can be coerced to its underlying value of -1 by other arithmetic operations.
Eg:
? True * 1 '>> -1
? False * 1 '>> 0
So this:
s = Int(xVal) + (xVal = n + 1)
is like writing:
If xVal = n + 1 Then
s = Int(xVal) + -1
else
s = Int(xVal) + 0
end if
or:
s = Int(xVal) + IIf(xVal = n + 1, -1, 0)
[(0,
'0.011*"people" + 0.009*"christian" + 0.008*"god" + 0.008*"law" + '
'0.006*"believe" + 0.005*"question" + 0.005*"man" + 0.005*"life" + '
'0.005*"time" + 0.005*"write"'),
(1,
'0.014*"organization" + 0.013*"group" + 0.012*"image" + 0.010*"university" + '
'0.009*"program" + 0.008*"newsletter" + 0.007*"graphic" + '
'0.007*"information" + 0.007*"file" + 0.006*"box"'),
(2,
'0.015*"write" + 0.015*"organization" + 0.014*"article" + 0.012*"year" + '
'0.008*"university" + 0.007*"team" + 0.007*"time" + 0.006*"game" + '
'0.006*"give" + 0.006*"kid"'),
(3,
'0.049*"space" + 0.009*"year" + 0.008*"publish" + 0.006*"aerospace" + '
'0.006*"news" + 0.006*"technical" + 0.005*"satellite" + 0.005*"activity" + '
'0.005*"membership" + 0.005*"system"')]
How do I set the delimeter for the text file shown in the image? I want it to split into four separate text files. What and how should I give the start and end delimiter in the if() as can be seen in the code?. The text file has four separate parts 0,1,2,3. I am trying to write all the parts into separate text file.
`with open('topics.txt','r') as fo:
op=''
start=0
cntr = 1
for x in fo.read().split("\n"):
if (x==''):
if (start==1):
with open(str(cntr) + '.txt','w') as opf:
opf.write(op)
opf.close()
op=''
cntr+=1
else:
start=1
else:
if (op==''):
op = x
else:
op = op + '\n' + x
fo.close()`
If what you posted above is literally your text file, then this should give you each tuple separately.
I'm just using the regular expressions library. The pattern is just look for a left paren ( and run of anything that isn't a right paren, and then a right paren. Super simple.
import re
foo = """[(0,\n blahblahblah), (1,\n asdfasdf), (2,\n ghhgghiegiegieh)]"""
pat = r'\([^\)]+\)'
matches = re.findall(pat, foo)
['(0,\n blahblahblah)', '(1,\n asdfasdf)', '(2,\n ghhgghiegiegieh)']
If you want to separate out the numbers you can do that easily by spliting and striping out the extra stuff:
[i[1:-1].split(',\n')[1].strip() for i in matches]
#['blahblahblah', 'asdfasdf', 'ghhgghiegiegieh']
Then you can write them to whatever file you like.
Trying to rotate an svg group by an amount deg in
data():{
deg: 90,
groupCenter: [100,200]
}
I have searched but can't find the correct syntax to bind data to the svg rotate function,
:transform="{'rotate(' + deg + ' ' + groupCenter[0] + ' ' + groupCenter[1] + ')'}"
I am trying to add this to a circle like this,
<circle style="mix-blend-mode: multiply;" v-for="(el,index) in element.coords" :fill="el.color" :key="index" :r="el.radius" :cy="el.y" :cx="el.x" :transform="{'rotate(' + deg + ' ' + groupCenter[0] + ' ' + groupCenter[1] + ')'}" />
I get,
- invalid expression: Unexpected token + in
{'rotate(' + deg + ' ' + groupCenter[0] + ' ' + groupCenter[1] + ')'}
Raw expression: :transform="{'rotate(' + deg + ' ' + groupCenter[0] + ' ' + groupCenter[1] + ')'}"
SOLUTION:
This worked for me using a method,
:transform="rotateShape(index)"
And the method,
rotateShape(){
return "rotate(" + this.deg + " 0 0)"
},
But I don't know why the initial attempt wouldn't work.
The transform attribute needs to evaluate to a string - i.e. transform="rotate(90deg)", so I think you need to concatenate some partial strings:
:transform="'rotate(' + deg + ' deg)'"
This is a great place to use javascript's template literals.
:transform="`rotate(${deg} deg)`"
I have created a function via vba and I have used this function to make an iterative table. I have set the workbook calculation to automatic and it all works fine but when I open the excel file, the cells that contain the mentioned function, give me #name error and everytime I need to recalculate. Is there a way to fix this?
Public Function FrictionFactor(relativeroughness, reynoldsnumber)
'Dim relativeroughness, reynoldsnumber As Double
fNext = 0.005 ' initial value for f
fIncrement = 0.005 ' initial step size
Convergence = 0.000001 ' sets the decimal place accuracy of the result
Do
fStart = fNext
LHSColebrookStart = 1 / (fStart ^ 0.5)
RHSColebrookStart = -2 * (Log((relativeroughness / 3.7) + (2.51 / (reynoldsnumber * (fStart ^ 0.5)))) / Log(10))
DifferenceStart = LHSColebrookStart - RHSColebrookStart
fNext = fStart + fIncrement
LHSColebrookNext = 1 / (fNext ^ 0.5)
RHSColebrookNext = -2 * (Log((relativeroughness / 3.7) + (2.51 / (reynoldsnumber * (fNext ^ 0.5)))) / Log(10))
DifferenceNext = LHSColebrookNext - RHSColebrookNext
If DifferenceStart * DifferenceNext < 0 Then ' march f in opposite direction and more slowly
fIncrement = fIncrement / -10
ElseIf DifferenceStart * DifferenceNext = 0 Then ' done
fIncrement = 0
End If ' keep marching f in same direction and at same rate
Loop While Abs(fStart - fNext) > Convergence
FrictionFactor = fStart
End Function
The usual reason this happens is that macros are not enabled when the workbook is opened. Check your Security settings.
This question already has answers here:
How to convert a column number (e.g. 127) into an Excel column (e.g. AA)
(60 answers)
Closed 8 years ago.
I need some help in doing a logic that would convert a numeric value to corresponding MS Excel header value.
For example:
1 = "A"
2 = "B"
3 = "C"
4 = "D"
5 = "E"
.........
25 = "Y"
26 = "Z"
27 = "AA"
28 = "AB"
29 = "AC"
30 = "AD"
.........
Would appreciate some .NET codes (C# or VB) for this. Thanks.
Here's some VBA (with test code) I strung together in Excel which does the trick. Unless VB.NET has changed drastically, it should work okay. Even if it has, you should be able to translate the idea into workable code.
' num2col - translate Excel column number (1-n) into column string ("A"-"ZZ"). '
Function num2col(num As Integer) As String
' Subtract one to make modulo/divide cleaner. '
num = num - 1
' Select return value based on invalid/one-char/two-char input. '
If num < 0 Or num >= 27 * 26 Then
' Return special sentinel value if out of range. '
num2col = "-"
Else
' Single char, just get the letter. '
If num < 26 Then
num2col = Chr(num + 65)
Else
' Double char, get letters based on integer divide and modulus. '
num2col = Chr(num \ 26 + 64) + Chr(num Mod 26 + 65)
End If
End If
End Function
' Test code in Excel VBA. '
Sub main()
MsgBox ("- should be " & num2col(0))
MsgBox ("A should be " & num2col(1))
MsgBox ("B should be " & num2col(2))
MsgBox ("Z should be " & num2col(26))
MsgBox ("AA should be " & num2col(27))
MsgBox ("AB should be " & num2col(28))
MsgBox ("AY should be " & num2col(51))
MsgBox ("AZ should be " & num2col(52))
MsgBox ("BA should be " & num2col(53))
MsgBox ("ZY should be " & num2col(27 * 26 - 1))
MsgBox ("ZZ should be " & num2col(27 * 26))
MsgBox ("- should be " & num2col(27 * 26 + 1))
End Sub
public string ColumnNumberToLetter(int ColumnNumber)
{
if (ColumnNumber > 26)
{
return ((char) (Math.Floor(((double)ColumnNumber - 1) / 26) + 64)).ToString()
+ ((char) (((ColumnNumber - 1) % 26) + 65)).ToString();
}
return ((char)(ColumnNumber+64)).ToString();
}
Try this:
public static string ToExcelString(int number)
{
if (number > 25)
{
int secondaryCounter = 0;
while (number > 25)
{
secondaryCounter = secondaryCounter + 1;
number = number - 25;
}
return ToExcelChar(number) + ToExcelChar(secondaryCounter);
}
else
{
return ToExcelChar(number)
}
}
private const string alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static string ToExcelChar(int number)
{
if (number > 25 || number < 0)
{
throw new InvalidArgumentException("the number passed in (" + number + ") must be between the range 0-25");
}
return alphabet[number];
}
Use a number base conversion routine. You want to convert from base 10 to base 26. Add each digit to 'A'
As in: http://www.vbforums.com/showthread.php?t=271359
Just Use the activecell.address then manuipulate the string based on where the $ is to what you need.