I want to validate the text within a textbox using a regular expression.
The text should be a number greater than 0 and less than and equal to 1000.
"^[1-9][0-9]*{1,2}$" is the regex you are looking for.
if(Regex.IsMatch(YourTextBox.Text,"^[1-9][0-9]*{1,2}$"))
{
//Write your logic here
}
Try this regex:
//for 0 < x < 1000
^((?<=[1-9])0|[1-9]){1,3}$
explain:
(?<=[1-9])0 //look behind to see if there is digits (1-9)
test:
0 -> invalid
000 -> invalid
45 -> valid
5 -> valid 'Ashwin Singh' solution can not capture this
101 -> valid
999 -> valid
1000 -> invalid
12345 -> invalid
10000 -> invalid
2558 -> invalid
205 -> valid
1001 -> invalid
2000 -> invalid
And better way convert to Decimal (if you dont use regular expression validator):
Decimal dc = Decimal.TryParse(textBox.Text);
if( dc > 0 && dc < 1000)
// do some thing
I found it:
^([1-9]|[1-9][0-9]|[1-9][0-9][0-9])$|^(1000)
I test it in range 0~1000
Related
I setup a constraint that does not constraint the solver in pyomo.
The constraint is the following:
def revenue_positive(model,t):
for t in model.T:
return (model.D[t] * model.P[t]) >= 0
model.positive_revenue = Constraint(model.T, rule=revenue_positive)
while the model parameters are:
model = ConcreteModel()
model.T = Set(doc='quarter of year', initialize=df.index.tolist(), ordered=True)
model.P = Param(model.T, initialize=df['price'].to_dict(), within=Any, doc='Price for each quarter')
model.C = Var(model.T, domain=NonNegativeReals)
model.D = Var(model.T, domain=NonNegativeReals)
income = sum(df.loc[t, 'price'] * model.D[t] for t in model.T)
expenses = sum(df.loc[t, 'price'] * model.C[t] for t in model.T)
profit = income - expenses
model.objective = Objective(expr=profit, sense=maximize)
# Solve the model
solver = SolverFactory('cbc')
solver.solve(model)
df dataframe is:
df time_stamp price Status imbalance Difference Situation ... week month hour_of_day day_of_week day_of_year yearly_quarter
quarter ...
0 2021-01-01 00:00:00 64.84 Final 16 -3 Deficit ... 00 1 0 4 1 1
1 2021-01-01 00:15:00 13.96 Final 38 2 Surplus ... 00 1 0 4 1 1
2 2021-01-01 00:30:00 12.40 Final 46 1 Surplus ... 00 1 0 4 1 1
3 2021-01-01 00:45:00 7.70 Final 65 14 Surplus ... 00 1 0 4 1 1
4 2021-01-01 01:00:00 64.25 Final 3 -9 Deficit ... 00 1 1 4 1 1
The objective is to constraint the solver not to accept a negative revenue. As such it does not work as the solver passes 6 negative revenue values through. Looking at the indices with negative revenue, it appears the system chooses to sell at a negative price to buy later at a price even "more" negative, so from an optimization standpoint, it is ok. I would like to check the difference in results if we prohibit the solver to do that. Any input is welcome as after many searches on the web, still not the right way to write it correctly.
I did a pprint() of the constraint that returned:
positive_revenue : Size=35040, Index=T, Active=True
UPDATE following new constraint code:
def revenue_positive(model,t):
return model.D[t] * model.P[t] >= 0
model.positive_revenue = Constraint(model.T, rule=revenue_positive)
Return the following error:
ERROR: Rule failed when generating expression for constraint positive_revenue
with index 283: ValueError: Invalid constraint expression. The constraint
expression resolved to a trivial Boolean (True) instead of a Pyomo object.
Please modify your rule to return Constraint.Feasible instead of True.
Error thrown for Constraint 'positive_revenue[283]'
ERROR: Constructing component 'positive_revenue' from data=None failed:
ValueError: Invalid constraint expression. The constraint expression
resolved to a trivial Boolean (True) instead of a Pyomo object. Please
modify your rule to return Constraint.Feasible instead of True.
Error thrown for Constraint 'positive_revenue[283]'
Traceback (most recent call last):
File "/home/olivier/Desktop/Elia - BESS/run_imbalance.py", line 25, in <module>
results_df = optimize_year(df)
File "/home/olivier/Desktop/Elia - BESS/battery_model_imbalance.py", line 122, in optimize_year
model.positive_revenue = Constraint(model.T, rule=revenue_positive)
File "/home/olivier/anaconda3/lib/python3.9/site-packages/pyomo/core/base/block.py", line 542, in __setattr__
self.add_component(name, val)
File "/home/olivier/anaconda3/lib/python3.9/site-packages/pyomo/core/base/block.py", line 1087, in add_component
val.construct(data)
File "/home/olivier/anaconda3/lib/python3.9/site-packages/pyomo/core/base/constraint.py", line 781, in construct
self._setitem_when_not_present(
File "/home/olivier/anaconda3/lib/python3.9/site-packages/pyomo/core/base/indexed_component.py", line 778, in _setitem_when_not_present
obj.set_value(value)
File "/home/olivier/anaconda3/lib/python3.9/site-packages/pyomo/core/base/constraint.py", line 506, in set_value
raise ValueError(
ValueError: Invalid constraint expression. The constraint expression resolved to a trivial Boolean (True) instead of a Pyomo object. Please modify your rule to return Constraint.Feasible instead of True.
Error thrown for Constraint 'positive_revenue[283]'
So there are 2 issues w/ your constraint. It isn't clear if one is a cut & paste issue or not.
The function call to make the constraint appears to be indented and inside of your function after the return statement, making it unreachable code. Could be just the spacing in your post.
You are incorrectly adding a loop inside of your function. You are passing in the parameter t as a function argument and then you are blowing it away with the for loop, which only executes for the first value of t in T then hits the return statement. Remove the loop. When you use the rule= structure in pyomo it will call the rule for each instance of the set that you are using in the Constraint(xx, rule=) structure.
So I think you should have:
def revenue_positive(model, t):
return model.D[t] * model.P[t] >= 0
model.positive_revenue = Constraint(model.T, rule=revenue_positive)
Updated re: the error you added.
The error cites the 283rd index. My bet is that price[283] is zero, so you are multiplying by a zero and killing your variable.
You could add a check within the function that checks if the price is zero, and in that case, just return pyo.Constraint.Feasible, which is the trivial return that doesn't influence the model (or crash)
I was given an encrypted copy of the study guide here, but how do you decrypt and read it???
In a file called pa11.py write a method called decode(inputfile,outputfile). Decode should take two parameters - both of which are strings. The first should be the name of an encoded file (either helloworld.txt or superdupertopsecretstudyguide.txt or yet another file that I might use to test your code). The second should be the name of a file that you will use as an output file.
Your method should read in the contents of the inputfile and, using the scheme described in the hints.txt file above, decode the hidden message, writing to the outputfile as it goes (or all at once when it is done depending on what you decide to use).
The penny math lecture is here.
"""
Program: pennyMath.py
Author: CS 1510
Description: Calculates the penny math value of a string.
"""
# Get the input string
original = input("Enter a string to get its cost in penny math: ")
cost = 0
Go through each character in the input string
for char in original:
value = ord(char) #ord() gives us the encoded number!
if char>="a" and char<="z":
cost = cost+(value-96) #offset the value of ord by 96
elif char>="A" and char<="Z":
cost = cost+(value-64) #offset the value of ord by 64
print("The cost of",original,"is",cost)
Another hint: Don't forget about while loops...
Another hint: After letters -
skip ahead by their pennymath value positions + 2
After numbers - skip ahead by their number + 7 positions
After anything else - just skip ahead by 1 position
The issue I'm having in that I cant seem to get the coding right to decode the file it comes out looking the same. This is the current code I have been using. But once I try to decrypt the message it stays the same.
def pennycost(c):
if c >="a" and c <="z":
return ord(c)-96
elif c>="A" and c<="Z":
return ord(c)-64
def decryption(inputfile,outputfile):
with open(inputfile) as f:
fo = open(outputfile,"w")
count = 0
while True:
c = f.read(1)
if not c:
break;
if count > 0:
count = count -1;
continue
elif c.isalpha():
count = pennycost(c)
fo.write(c)
elif c.isdigit():
count = int(c)
fo.write(c)
else:
count = 6
fo.write(c)
fo.close()
inputfile = input("Please enter the input file name: ")
outputfile = input("Plese enter the output file name(EXISTING FILE WILL BE OVER WRITTEN!): ")
decryption(inputfile,outputfile)
I am trying to make a program to prompt a user for input until they enter a number within a specific range.
What is the best approach to make sure the code does not error out when I enter a letter, a symbol, or a number outside of the specified range?
In alternative to parse, you can use tryparse:
tryparse(type, str; base)
Like parse, but returns either a value of the requested type, or
nothing if the string does not contain a valid number.
The advantage over parse is that you can have a cleaner error handling without resorting to try/catch, which would hide all exceptions raised within the block.
For example you can do:
while true
print("Please enter a whole number between 1 and 5: ")
input = readline(stdin)
value = tryparse(Int, input)
if value !== nothing && 1 <= value <= 5
println("You entered $(input)")
break
else
#warn "Enter a whole number between 1 and 5"
end
end
Sample run:
Please enter a whole number between 1 and 5: 42
┌ Warning: Enter a whole number between 1 and 5
└ # Main myscript.jl:9
Please enter a whole number between 1 and 5: abcde
┌ Warning: Enter a whole number between 1 and 5
└ # Main myscript.jl:9
Please enter a whole number between 1 and 5: 3
You entered 3
This is one possible way to achieve this sort of thing:
while true
print("Please enter a whole number between 1 and 5: ")
input = readline(stdin)
try
if parse(Int, input) <= 5 || parse(Int, input) >= 1
print("You entered $(input)")
break
end
catch
#warn "Enter a whole number between 1 and 5"
end
end
Sample Run:
Please enter a whole number between 1 and 5: 2
You entered 2
See this link for how to parse the user input into an int.
I'm using an Hash Table to store some values. Here are the details:
There will be roughly 1M items to store (not known before, so no perfect-hash possible).
Table is 10M large.
Hash function is MurMurHash3.
I did some tests and storing 1M values I get 350,000 collisions and 30 elements at the most-colliding hash table's slot.
Are these result good?
Would it make sense to implement Binary Search for lists that get created at colliding hash-table's slots?
What' your advice to improve performances?
EDIT: Here is my code
var
HashList: array [0..10000000 - 1] of Integer;
for I := 0 to High(HashList) do
HashList[I] := 0;
for I := 1 to 1000000 do
begin
Y := MurmurHash3(UIntToStr(I));
Y := Y mod Length(HashList);
Inc(HashList[Y]);
if HashList[Y] > 1 then
Inc(TotalCollisionsCount);
if HashList[Y] > MostCollidingSlotItemCount then
MostCollidingSlotItemCount := HashList[Y];
end;
Writeln('Total: ' + IntToStr(TotalCollisionsCount) + ' Max: ' + IntToStr(MostCollidingSlotItemCount));
Here is the result I get:
Total: 48169 Max: 5
Am I missing something?
This is what you get when you put 1M items randomly into 10M cells
calendar_size=10000000 nperson = 1000000
E/cell| Ncell | frac | Nelem | frac |h/cell| hops | Cumhops
----+---------+--------+----------+--------+------+--------+--------
0: 9048262 (0.904826) 0 (0.000000) 0 0 0
1: 905064 (0.090506) 905064 (0.905064) 1 905064 905064
2: 45136 (0.004514) 90272 (0.090272) 3 135408 1040472
3: 1488 (0.000149) 4464 (0.004464) 6 8928 1049400
4: 50 (0.000005) 200 (0.000200) 10 500 1049900
----+---------+--------+----------+--------+------+--------+--------
5: 10000000 1000000 1.049900 1049900
The left column is the number of items in a cell. The second: the number of cells having this itemcount.
WRT the binary search: it is obvious that for small tables like this (maximum chain length=4, but most chains are of length=1), linear search outperforms binary search. The takeover-point is probably somewhere between 10 and 100.
I would like to remove the duplicate records from my large .xdf file trans.xdf.
Here is the file details:
File name: /poc/revor/data/trans.xdf
Number of observations: 1000000000
Number of variables: 5
Number of blocks: 40
Compression type: zlib
Variable information:
Var 1: CARD_ID, Type: character
Var 2: SE_NO, Type: character
Var 3: r12m_cv, Type: numeric, Low/High: (-2348.7600, 40587.3900)
Var 4: r12m_roc, Type: numeric, Low/High: (0.0000, 231.0000)
Var 5: PROD_GRP_CD, Type: character
Also below is the sample data of the file:
CARD_ID SE_NO r12m_cv r12m_roc PROD_GRP_CD
900000999000000000 1045815024 110 1 1
900000999000000000 1052487253 247.52 2 1
900000999000000000 9999999999 38.72 1 1
900000999000000000 1090389768 1679.96 16 1
900000999000000000 1091226035 0 1 1
900000999000000000 1091241208 538.68 4 1
900000999000000000 9999999999 83 1 1
900000999000000000 1091468041 148.4 3 1
900000999000000000 1092640358 3.13 1 1
900000999000000000 1093468692 546.29 1 1
I have tried using rxDataStep function to use its transform parameter to call to unique() function over the .xdf file. Below is the code for the same:
uniq_dat <- function( dataList )
{
datalist <- unique(datalist)
return(datalist)
}
rxDataStepXdf(inFile = "/poc/revor/data/trans.xdf",outFile = "/poc/revor/data/trans.xdf",transformFunc = uniq_dat,overwrite = TRUE)
But was getting below error:
Error in unique(datalist) : object 'datalist' not found
Error in transformation function: Error in unique(datalist) : object 'datalist' not found
Error in rxCall("RxDataStep", params) :
So anybody could point out the mistake that I am doing here or if there is a better way to remove the duplicate records from the .Xdf file. I am avoiding loading the data into inmemory dataframe as the data is pretty huge.
I am running the above code in Revolution R Environment over HDFS.
If the same can be obtained by any other approach then the example for the same would be appreciated.
Thanks for the help in advance :)
Cheers,
Amit
you can remove the duplicate values providing removeDupKeys=TRUE parameter for rxSort() function. For example for your case:
XdfFilePath <- file.path("<your file's fully qualified path>/trans.xdf")
rxSort(inData = XdfFilePath,sortByVars=c("CARD_ID","SE_NO","r12m_cv","r12m_roc","PROD_GRP_CD"), removeDupKeys=TRUE)
if you want to remove duplicate records based on a specific key column, for example, based on SE_NO column
set the key value as sortByVars="SE_NO"