i <- 1
while(i<10){
print(i)
i<-i+1
if(i==4){
next
}
}
[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
[1] 6
[1] 7
[1] 8
[1] 9
I execute this code, but still have number 4 in my result although I am using "next" in my code to skipp it
use the following instead.
if(i!=4){print(i)}
or you wanted to test with next, use print after next
{
if(i==4){next}
print(i)
}
Related
My issue is, i don't know how to use the output of a function properly. The output contains multiple lines (j = column , i = testresult)
I want to use the output for some other rules in other functions. (eg. if (i) testresult > 5 then something)
I have a function with two loops. The function goes threw every column and test something. This works fine.
def test():
scope = range(10)
scope2 = range(len(df1.columns))
for (j) in scope2:
for (i) in scope:
if df1.iloc[:,[j]].shift(i).loc[selected_week].item() > df1.iloc[:,[j]].shift(i+1).loc[selected_week].item():
i + 1
else:
print(j,i)
break
Output:
test()
1 0
2 3
3 3
4 1
5 0
6 6
7 0
8 1
9 0
10 1
11 1
12 0
13 0
14 0
15 0
I tried to convert it to list, dataframe etc. However, i miss something here.
What is the best way for that?
Thank you!
A fix of your code would be:
def test():
out = []
scope = range(10)
scope2 = range(len(df1.columns))
for j in scope2:
for i in scope:
if df1.iloc[:,[j]].shift(i).loc[selected_week].item() <= df1.iloc[:,[j]].shift(i+1).loc[selected_week].item():
out.append([i, j])
return pd.DataFrame(out)
out = test()
But you probably don't want to use loops as it's slow, please clarify what is your input with a minimal reproducible example and what you are trying to achieve (expected output and logic), we can probably make it a vectorized solution.
I want to modify each element of a block by foreach. I tried like this but failed:
>> a: [3 4 5 6]
== [3 4 5 6]
>> foreach i a [i + 1]
== 7
>> a
== [3 4 5 6] ;; block a is not changed. What I want is [4 5 6 7]
Is there a better way to achieve it?
Changes that you made to values do not persist in a block itself. This ties back to your question about call-by-value parameter passing in Rebol and Red: you modify a copy on the stack (passed down to + along with 1), not the actual value slot that sits inside block a.
To achieve what you want, you need to increment integers in-place, without pushing them on the stack. One way to do so is by using forall.
>> block: [1 2 3]
== [1 2 3]
>> also block forall block [block/1: block/1 + 1]
== [2 3 4]
What forall does is setting a word to a series and then incrementally bumping its index:
>> forall block [probe block]
[1 2 3]
[2 3]
[3]
Since it doesn't extract the actual values, you can access them using path notation, and then modify them in place. block/1 always pick the first value on each iteration.
As usual, no reply with your foreach.
a: [2 3 4 5]
b: copy []
foreach i a [append b i + 1]
and if you wish you can set a to b now
a: b
The problem with doing this in one step is that you do not have an index you can use here (despite the suggestive letter i, but that is representing the content of each item inside the block).
So now you can choose your favourite solution.
>> help forall
USAGE:
FORALL 'word body
DESCRIPTION:
Evaluates body for all values in a series.
FORALL is a native! value.
ARGUMENTS:
'word [word!] "Word referring to series to iterate over."
body [block!]
use forall
> a: [3 4 5 6]
== [3 4 5 6]
>> forall a [a/1: a/1 + 1]
== 7
>> probe a
[4 5 6 7]
== [4 5 6 7]
I am trying to fill each row in a new column (Previous time) with a value from previous row of the specific subset (when condition is met). The thing is, that if I interrupt kernel and check values, it is ok. But if it runs to the end, then all rows in new column are filled with None. If previous row doesnt exist, than I will fill it with first value.
Name First round Previous time
Runner 1 2 2
Runner 2 5 5
Runner 3 5 5
Runner 1 6 2
Runner 2 8 5
Runner 3 4 5
Runner 1 2 6
Runner 2 5 8
Runner 3 5 4
What I tried:
df.insert(column = "Previous time", value = 999)
def fce(arg):
runner= arg[0]
stat = arg[1]
if stat == 999:
# I used this to avoid filling all rows in a new column again for the same runner
first = df.loc[df['Name'] == runner,"First round"].iloc[0]
df.loc[df['Name'] == runner,"Previous time"] = df.loc[df['Name'] == runner]["First round"].shift(1, fill_value = first)
df["Previous time"] = df[['Name', "Previous time"]].apply(fce, axis=1)
Condut gruopby shift for each Name and fill the missing values with the original series.
df['Previous time'] = (df.groupby('Name')['First round']
.shift()
.fillna(df['First round'], downcast='infer'))
The problem is that your function fce returns None for every row, so the Series produced by the term df[['Name', "Previous time"]].apply(fce, axis=1) is a Series of None.
That is, instead of overriding the Dataframe with df.loc inside the function, you need to return the value to fill for this position. Unfortunately, this is impossible since then you need to know which indices you already calculated.
A better way to do it would be to use groupby. This is a more natural way, since you want to perform an action on each group. If you use apply after groupby and you to return a series, you, in fact, define a value for each row. Just remember to remove the extra index "Name" that groupby adds.
def fce(g):
first = g["First round"].iloc[0]
return g["First round"].shift(1, fill_value=first)
df["Previous time"] == df.groupby("Name").apply(fce).reset_index("Name", drop=True)
Thank you very much. Please can you answer me one more question? How does it work with group by on multiple columns if I want to return mean of all rounds based on specific runner a sleeping time before race.
Expected output:
Name First round Sleep before race Mean
Runner 1 2 8 4
Runner 2 5 7 6
Runner 3 5 8 5
Runner 1 6 8 4
Runner 2 8 7 6
Runner 3 4 9 4,5
Runner 1 2 9 2
Runner 2 5 7 6
Runner 3 5 9 4,5
This does not work for me.
def last_season(g):
aa = g["First round"].mean()
df["Mean"] = df.groupby(["Name", "Sleep before race"]).apply(g).reset_index(["Name", "Sleep before race"], drop=True)
I'm trying to use sqlSave command to import R dataframe into SQL database. Below is my code
> head(final_series)
Price Time FactorID CountryID id
1 5.363334e+01 1980-01-01 1 1 1
2 5.143333e+01 1980-04-01 1 1 16384
3 5.060000e+01 1980-07-01 1 1 32767
4 5.250000e+01 1980-10-01 1 1 49150
5 5.266667e+01 1981-01-01 1 1 65533
6 5.280000e+01 1981-04-01 1 1 81916
> sqlSave(dbhandle, final_series, tablename = "db_time_price", varTypes = c(id="uniqueidentifier", FactorID= "float", CountryID="float", Time="date", Price="float"), append=TRUE, verbose = T, fast = F)
But I got the following error:
Error in dimnames(x) <- dn :
length of 'dimnames' [2] not equal to array extent
Anyone knows why? Thanks!
Did you check if the table already exists? If the table already exists but with a different dimension you would see this error.
I am using Gurobi 7 to solve my MIP. I have several different variables. However, I am specifically interested in two of those, "x" and "y" namely. For the reference, I am giving my code that shows how I added x and y variables into the solver:
# Creating Variables
x = {}
y = {}
# Adding Variables
for i in range(I):
x[i+1,P[i]-d[0]] = m.addVar(vtype=GRB.BINARY, name="x%s" % str([i+1,P[i]-d[0]]))
x[i+1,P[i]] = m.addVar(vtype=GRB.BINARY, name="x%s" % str([i+1,P[i]]))
for i in range(I):
for k in range(len(rangevalue)):
y[i+1, rangevalue[k] - E[i]] = m.addVar(vtype=GRB.BINARY,
name="y%s" % str([i+1, rangevalue[k] - E[i]]))
Even though the above code may not really make any sense, I just wanted to show it in case you may use it for my problem.
After I solve the problem, I get the following results:
m.printAttr('X')
Variable X
-------------------------
x[1, 3] 1
sigmaminus[1] 874
x[2, 2] 1
sigmaminus[2] 1010
x[3, 2] 1
sigmaminus[3] 1945
x[4, 4] 1
sigmaplus[4] 75
x[5, 4] 1
sigmaminus[5] 1153
x[6, 5] 1
sigmaminus[6] 280
x[7, 3] 1
sigmaplus[7] 1138
x[8, 2] 1
sigmaplus[8] 538
x[9, 1] 1
sigmaplus[9] 2432
x[10, 5] 1
sigmaminus[10] 480
omega[1] 12
OMEGA[1] 12
omega[2] 9
OMEGA[2] 12
omega[3] 8
OMEGA[3] 9
omega[4] 8
OMEGA[4] 8
OMEGA[5] 8
y[1, 2] 1
y[2, 9] 1
y[3, 5] 1
y[4, 6] 1
y[5, 4] 1
y[6, 6] 1
y[7, 3] 1
y[8, 11] 1
y[9, 8] 1
y[10, 1] 1
phiplus[6] 1
phiminus[7] 1
phiminus[10] 1
I specifically want to display x and y variables with their indexes. Other variables are not necessary. My question is how can I write these results into an csv file on one column as following?
x[1,3]
x[2,2]
x[3,2]
.
.
.
x[10,5]
y[1,2]
y[2,9]
y[3,5]
.
.
.
y[10,1]
I do not need their corresponding value which can only be "1" since they are binary variables. I just need to write the variables which have the value "1".
I would do something along these lines:
import csv
if m.SolCount == 0:
print("Model has no solution")
exit(1)
var_names = []
for var in m.getVars():
# Or use list comprehensions instead
if 'x' == str(var.VarName[0]) and var.X > 0.1:
var_names.append(var.VarName)
# Write to csv
with open('out.csv', 'wb') as myfile:
wr = csv.writer(myfile, quoting=csv.QUOTE_ALL)
wr.writerows(var_names)
I hope this helps. I am going to test this snippet a bit later. Update: works as intended.