Process by rows adding values - excel

I'm trying to transpose and sum with the following criteria: I have to create a row for each LOGIN and DATE and a column with the ACT values and the sum of their respective MAP values. In the middle separated by : I have to create the sum of all the MAP values, as follows:
LOGIN DATE ACT MAP
1 11/02/2008 149 3
1 11/02/2008 18 1
1 11/02/2008 18 1
1 11/02/2008 18 5
1 13/02/2008 145 2
1 13/02/2008 43 3
2 13/02/2008 19 0
2 13/02/2008 18 1
2 14/02/2008 18 1
2 14/02/2008 18 1
3 14/02/2008 39 1
3 15/02/2008 149 0
3 15/02/2008 43 0
3 15/02/2008 19 1
3 15/02/2008 19 1
1 11/02/2008 149 18 : 10: 3 7 This is the first row that I should create because 149 and 18 are the ACT values for this LOGIN and DATE, 3 = MAP value for ACT 149 and 7 is the sum of the MAP values for ACT 18, 7=1+1+5, in the middle the 10 value = 3+7
1 13/02/2008 145 43 : 5: 2 3
2 13/02/2008 19 18 : 1: 1 0
2 14/02/2008 18 : 2 : 2
3 14/02/2008 39 : 1 : 1
3 15/02/2008 149 43 19 : 2 : 0 0 2
I grouped and added to obtain this but need to process by rows
LOGIN MAP
1 15
11/02/2008 10
13/02/2008 5
2 3
13/02/2008 1
14/02/2008 2
3 3
14/02/2008 1
15/02/2008 2

I transformed the input file and now it looks like this, now I need to concatenate the values of the ACT column until I find a blank row. For example I need to create 18 149 10 7 3 for the first group until the first blank. For the second blank I need to create 43 145 5 3 2
LOGIN ACT Total
1 18 7
1 149 3
1 10
1 43 3
1 145 2
1 5
2 18 1
2 19 0
2 1
2 18 2
2 2
3 39 1
3 1
3 19 2
3 43 0
3 149 0
3 2

Related

how to shift column labels to left python

I have dataframe i want to move column name to left from specific column. original dataframe have many columns can not do this by rename columns
df=pd.DataFrame({'A':[1,3,4,7,8,11,1,15,20,15,16,87],
'H':[1,3,4,7,8,11,1,15,78,15,16,87],
'N':[1,3,4,98,8,11,1,15,20,15,16,87],
'p':[1,3,4,9,8,11,1,15,20,15,16,87],
'B':[1,3,4,6,8,11,1,19,20,15,16,87],
'y':[0,0,0,0,1,1,1,0,0,0,0,0]})
print((df))
A H N p B y
0 1 1 1 1 1 0
1 3 3 3 3 3 0
2 4 4 4 4 4 0
3 7 7 98 9 6 0
4 8 8 8 8 8 1
5 11 11 11 11 11 1
6 1 1 1 1 1 1
7 15 15 15 15 19 0
8 20 78 20 20 20 0
9 15 15 15 15 15 0
10 16 16 16 16 16 0
11 87 87 87 87 87 0
Here i want to remove label N first dataframe after removing label N
A H p B y
0 1 1 1 1 1 0
1 3 3 3 3 3 0
2 4 4 4 4 4 0
3 7 7 98 9 6 0
4 8 8 8 8 8 1
5 11 11 11 11 11 1
6 1 1 1 1 1 1
7 15 15 15 15 19 0
8 20 78 20 20 20 0
9 15 15 15 15 15 0
10 16 16 16 16 16 0
11 87 87 87 87 87 0
Rrquired output:
A H P B y
0 1 1 1 1 1 0
1 3 3 3 3 3 0
2 4 4 4 4 4 0
3 7 7 98 9 6 0
4 8 8 8 8 8 1
5 11 11 11 11 11 1
6 1 1 1 1 1 1
7 15 15 15 15 19 0
8 20 78 20 20 20 0
9 15 15 15 15 15 0
10 16 16 16 16 16 0
11 87 87 87 87 87 0
Here last column can be ignore
Note: in original dataframe have many columns , can not rename columns , so need some auto method to shift column names lef
You can do
df.columns=sorted(df.columns.str.replace('N',''),key=lambda x : x=='')
df
A H p B y
0 1 1 1 1 1 0
1 3 3 3 3 3 0
2 4 4 4 4 4 0
3 7 7 98 9 6 0
4 8 8 8 8 8 1
5 11 11 11 11 11 1
6 1 1 1 1 1 1
7 15 15 15 15 19 0
8 20 78 20 20 20 0
9 15 15 15 15 15 0
10 16 16 16 16 16 0
11 87 87 87 87 87 0
Replace the columns with your own custom list.
>>> cols = list(df.columns)
>>> cols.remove('N')
>>> df.columns = cols + ['']
Output
>>> df
A H p B y
0 1 1 1 1 1 0
1 3 3 3 3 3 0
2 4 4 4 4 4 0
3 7 7 98 9 6 0
4 8 8 8 8 8 1
5 11 11 11 11 11 1
6 1 1 1 1 1 1
7 15 15 15 15 19 0
8 20 78 20 20 20 0
9 15 15 15 15 15 0
10 16 16 16 16 16 0
11 87 87 87 87 87 0

How to save rows when value change in column python

I have DataFrame with two columns ID and Value1, I want to select rows when the value of column value1 column changes. I want to save rows 3 before change and 3 after the change and also change point row.
df=pd.DataFrame({'ID':[1,3,4,6,7,8,90,23,56,78,90,34,56,78,89,34,56],'Value1':[0,0,0,0,0,2,2,2,2,0,0,0,1,1,1,1,1]})
ID Value1
0 1 0
1 3 0
2 4 0
3 6 0
4 7 0
5 8 2
6 90 2
7 23 2
8 56 2
9 78 0
10 90 0
11 34 0
12 56 1
13 78 1
14 89 1
15 34 1
16 56 1
output:
ID Value1
0 4 0
1 6 0
2 7 0
3 8 2
4 90 2
5 23 2
6 90 2
7 23 2
8 56 2
9 78 0
10 90 0
11 34 0
IIUC,
import numpy as np
df=pd.DataFrame({'ID':[1,3,4,6,7,8,90,23,56,78,90,34,56,78,89,34,56],'Value1':[0,0,0,0,0,2,2,2,2,0,0,0,1,1,1,1,1]})
df.reset_index(drop=True) #index needs to start from zero for solution
ind = list(set([val for i in df[df['Value1'].diff()!=0].index for val in range(i-3, i+4) if i>0 and val>=0]))
# diff gives column wise differencing. combined it with nested list and
# finally, list(set()) to drop any duplicates in index values
df[df.index.isin(ind)]
ID Value1
2 4 0
3 6 0
4 7 0
5 8 2
6 90 2
7 23 2
8 56 2
9 78 0
10 90 0
11 34 0
12 56 1
13 78 1
14 89 1
15 34 1
If you want to retain occurrences of duplicates, drop the list(set()) function over the list

array index is greater than array upper bound for d

Can anyone help me with this issue please? I have used other data sets with different number of studies (NS) and treatments (NT) and it worked fine.
Any help will be highly appreciated.
The dataset is as follows:
list(N=186, NS=5, NT=3, mean=c(0,0);
Where
N= number of intervals
NS= number of studies
NT= number of treatment
s[]= Study ID
r[]= no of events
n[]= no at risk
t[]= study arm ID
b[]= Study arm base
time[]= time in months
dt[]= difference in interval (months)
model {
for (i in 1:N) { # N=number of datapoints in dataset
#likelihood
r[i] ~dbin(p[i],n[i])
p[i]<-1- exp(-h[i]*dt[i]) # hazard h over interval [t,t+dt] expressed as deaths per unit person-time (e.g. months)
#fixed effects model
log(h[i]) <- nu[i]+log(time[i])*theta[i]
nu[i]<-mu[s[i],1]+d[s[i],1]*(1-equals(t[i],b[i]))
theta[i]<-mu[s[i],2]+ d[s[i],2]*(1-equals(t[i],b[i]))
}
# priors
d[1,1]<- 0
d[1,2]<- 0
for(j in 2 :NT){ # NT=number of treatments
d[j,1:2] ~ dmnorm(mean[1:2],prec2[,])
}
for(k in 1:NS) {
mu[k,1:2] ~ dmnorm(mean[1:2],prec2[,])
}
}
#Winbugs data set
list(N=176, NS=5, NT=3, mean=c(0,0),
prec2 = structure(.Data = c(0.0001,0,0,0.0001), .Dim = c(2,2)))
# initials 1
list(
d=structure(.Data=c(NA,NA,0,0,0,0,0,0), .Dim = c(4,2)),
mu = structure(.Data=c(1,1,1,1,1,1,1,1), .Dim = c(4,2)))
# initials 2
list(
d=structure(.Data=c(NA,NA,0.5,0.5,0.5,0.5,0.5,0.5), .Dim = c(4,2)),
mu = structure(.Data=c(0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5), .Dim = c(4,2)))
s[] r[] n[] t[] b[] time[] dt[]
1 1 62 1 1 3 2
1 2 59 1 1 7 4
1 6 53 1 1 11 2
1 2 51 1 1 13 2
1 3 48 1 1 15 2
1 2 45 1 1 17 2
1 5 40 1 1 19 2
1 2 37 1 1 23 4
1 2 35 1 1 25 2
1 2 32 1 1 27 2
1 1 31 1 1 29 2
1 2 28 1 1 31 2
1 2 26 1 1 33 2
1 2 23 1 1 35 2
1 1 21 1 1 39 4
1 1 14 1 1 51 12
1 2 55 2 1 5 4
1 1 54 2 1 7 2
1 2 52 2 1 9 2
1 1 51 2 1 11 2
1 5 46 2 1 13 2
1 2 44 2 1 15 2
1 3 41 2 1 17 2
1 3 37 2 1 19 2
1 2 35 2 1 21 2
1 1 34 2 1 23 2
1 1 33 2 1 25 2
1 1 32 2 1 31 6
1 3 29 2 1 33 2
1 1 28 2 1 35 2
1 1 26 2 1 39 4
1 1 24 2 1 41 2
1 1 22 2 1 43 2
1 2 19 2 1 45 2
2 8 169 1 1 3 4
2 10 148 1 1 5 2
2 8 137 1 1 7 2
2 6 127 1 1 9 2
2 8 118 1 1 11 2
2 7 109 1 1 13 2
2 3 105 1 1 15 2
2 4 95 1 1 17 2
2 3 84 1 1 19 2
2 3 76 1 1 21 2
2 4 68 1 1 23 2
2 4 60 1 1 25 2
2 4 50 1 1 27 2
2 1 35 1 1 31 4
2 2 29 1 1 33 2
2 1 25 1 1 35 2
2 3 21 1 1 37 2
2 1 18 1 1 39 2
2 2 11 1 1 43 4
2 1 180 2 1 1 2
2 11 162 2 1 3 2
2 9 147 2 1 5 2
2 9 135 2 1 7 2
2 6 125 2 1 9 2
2 6 116 2 1 11 2
2 6 106 2 1 13 2
2 7 95 2 1 15 2
2 1 92 2 1 17 2
2 5 84 2 1 19 2
2 3 77 2 1 21 2
2 2 67 2 1 23 2
2 1 59 2 1 25 2
2 4 49 2 1 27 2
2 1 40 2 1 29 2
2 2 34 2 1 31 2
2 3 23 2 1 37 6
2 1 19 2 1 39 2
4 1 62 1 1 3 2
4 2 59 1 1 7 4
4 6 53 1 1 11 2
4 2 51 1 1 13 2
4 3 48 1 1 15 2
4 2 45 1 1 17 2
4 5 40 1 1 19 2
4 2 37 1 1 23 4
4 2 35 1 1 25 2
4 2 32 1 1 27 2
4 1 31 1 1 29 2
4 2 28 1 1 31 2
4 2 26 1 1 33 2
4 2 23 1 1 35 2
4 1 21 1 1 39 4
4 1 14 1 1 51 12
4 2 55 2 1 5 4
4 1 54 2 1 7 2
4 2 52 2 1 9 2
4 1 51 2 1 11 2
4 5 46 2 1 13 2
4 2 44 2 1 15 2
4 3 41 2 1 17 2
4 3 37 2 1 19 2
4 2 35 2 1 21 2
4 1 34 2 1 23 2
4 1 33 2 1 25 2
4 1 32 2 1 31 6
4 3 29 2 1 33 2
4 1 28 2 1 35 2
4 1 26 2 1 39 4
4 1 24 2 1 41 2
4 1 22 2 1 43 2
4 2 19 2 1 45 2
5 8 169 1 1 3 4
5 10 148 1 1 5 2
5 8 137 1 1 7 2
5 6 127 1 1 9 2
5 8 118 1 1 11 2
5 7 109 1 1 13 2
5 3 105 1 1 15 2
5 4 95 1 1 17 2
5 3 84 1 1 19 2
5 3 76 1 1 21 2
5 4 68 1 1 23 2
5 4 60 1 1 25 2
5 4 50 1 1 27 2
5 1 35 1 1 31 4
5 2 29 1 1 33 2
5 1 25 1 1 35 2
5 3 21 1 1 37 2
5 1 18 1 1 39 2
5 2 11 1 1 43 4
5 1 180 2 1 1 2
5 11 162 2 1 3 2
5 9 147 2 1 5 2
5 9 135 2 1 7 2
5 6 125 2 1 9 2
5 6 116 2 1 11 2
5 6 106 2 1 13 2
5 7 95 2 1 15 2
5 1 92 2 1 17 2
5 5 84 2 1 19 2
5 3 77 2 1 21 2
5 2 67 2 1 23 2
5 1 59 2 1 25 2
5 4 49 2 1 27 2
5 1 40 2 1 29 2
5 2 34 2 1 31 2
5 3 23 2 1 37 6
5 1 19 2 1 39 2
3 2 179 1 1 1 2
3 4 172 1 1 3 2
3 3 168 1 1 5 2
3 6 157 1 1 7 2
3 4 151 1 1 9 2
3 9 142 1 1 11 2
3 10 130 1 1 13 2
3 7 123 1 1 15 2
3 3 119 1 1 17 2
3 5 112 1 1 19 2
3 3 108 1 1 21 2
3 3 103 1 1 23 2
3 12 91 1 1 25 2
3 2 68 1 1 27 2
3 2 46 1 1 29 2
3 8 29 1 1 31 2
3 2 23 1 1 33 2
3 3 8 1 1 35 2
3 5 175 3 1 3 4
3 7 163 3 1 5 2
3 12 151 3 1 7 2
3 12 139 3 1 9 2
3 4 132 3 1 11 2
3 9 122 3 1 13 2
3 7 114 3 1 15 2
3 4 108 3 1 17 2
3 7 101 3 1 19 2
3 5 96 3 1 21 2
3 7 89 3 1 23 2
3 2 87 3 1 25 2
3 4 68 3 1 27 2
3 4 50 3 1 29 2
3 3 40 3 1 31 2
3 3 22 3 1 33 2
3 1 8 3 1 35 2
END
You have set NT = 3 while the indexed s vector ranges from 1 to 5.
Set NT = 5 or NT = length(unique(s)).

How to write Python code that does cumprod for forward 2 periods with groupby

I want to calculate Return, RET, which is Cumulative of 2 periods (now & next period) with groupby(id).
df['RET'] = df.groupby('id')['trt1m1'].rolling(2,min_periods=2).apply(lambda x:x.prod()).reset_index(0,drop=True)
Expected Result:
id datadate trt1m1 RET
1 20051231 1 2
1 20060131 2 6
1 20060228 3 12
1 20060331 4 16
1 20060430 4 20
1 20060531 5 Nan
2 20061031 10 110
2 20061130 11 165
2 20061231 15 300
2 20070131 20 420
2 20070228 21 Nan
Actual Result:
id datadate trt1m1 RET
1 20051231 1 Nan
1 20060131 2 2
1 20060228 3 6
1 20060331 4 12
1 20060430 4 16
1 20060531 5 20
2 20061031 10 Nan
2 20061130 11 110
2 20061231 15 165
2 20070131 20 300
2 20070228 21 420
The code i used calculate cumprod for trailing 2 periods instead of forward.

Getting a number of quarter from numeric week number and the week number within the quarter in python?

I've a list of number from 1 to 53. I am trying to calculate 1) the quarter of a week and 2) the number of that week within that quarter using numeric week numbers. (if 53, needs to be qtr 4 wk 14, if 27 needs to be 3rd quarter wk 1). Got this working in excel, but not in python? Any thoughts?
tried the following, but at each try I've an issue with the wk's like 13 or 27 depending on the method I'm using.
13 -> should be qtr 1 , 27 -> should be 3 qtr.
df['qtr1'] = df['wk']//13
df['qtr2']=(np.maximum((df['wk']-1),1)/13)+1
df['qtr3']=((df1['wk']-1)//13)
df['qtr4'] = df['qtr2'].astype(int)
Results are awkward
wk qtr qtr2 qtr3 qtr4
1.0 0 1.076923 -1.0 1
13.0 1(wrong) 1.923077 0.0 1
14.0 1 2.000000 1.0 2
27.0 2 3.000000 1.0 2 (wrong)
28.0 2 3.076923 2.0 3
You can convert your weeks to integers, by using astype:
df['wk'] = df['wk'].astype(int)
You should subtract it with one first, like:
df['qtr'] = ((df['wk']-1) // 13) + 1
df['weekinqtr'] = (df['wk']-1) % 13 + 1
since 13//13 will be 1, not zero. This gives us:
>>> df
wk qtr weekinqtr
0 1 1 1
1 13 1 13
2 14 2 1
3 26 2 13
4 27 3 1
5 28 3 2
If you want extra columns per quarter, you can use get_dummies(..) [pandas-doc] to obtain a one-hot encoding per quarter:
>>> df.join(pd.get_dummies(df['qtr'], prefix='qtr'))
wk qtr weekinqtr qtr_1 qtr_2 qtr_3
0 1 1 1 1 0 0
1 13 1 13 1 0 0
2 14 2 1 0 1 0
3 26 2 13 0 1 0
4 27 3 1 0 0 1
5 28 3 2 0 0 1
Using div // and modulo % work for what you want I think
In [254]: df = pd.DataFrame({'week':range(52)})
In [255]: df['qtr'] = (df['week'] // 13) + 1
In [256]: df['qtr_week'] = df['week'] % 13
In [257]: df.loc[(df['qtr_week'] ==0),'qtr_week']=13
In [258]: df
Out[258]:
week qtr qtr_week
0 1 1 1
1 2 1 2
2 3 1 3
3 4 1 4
4 5 1 5
5 6 1 6
6 7 1 7
7 8 1 8
8 9 1 9
9 10 1 10
10 11 1 11
11 12 1 12
12 13 2 13
13 14 2 1
14 15 2 2
15 16 2 3
16 17 2 4
17 18 2 5
18 19 2 6
19 20 2 7
20 21 2 8
21 22 2 9
22 23 2 10
23 24 2 11
24 25 2 12
25 26 3 13
26 27 3 1
27 28 3 2
28 29 3 3
29 30 3 4
30 31 3 5
31 32 3 6
32 33 3 7
33 34 3 8
34 35 3 9
35 36 3 10
36 37 3 11
37 38 3 12
38 39 4 13
39 40 4 1
40 41 4 2
41 42 4 3
42 43 4 4
43 44 4 5
44 45 4 6
45 46 4 7
46 47 4 8
47 48 4 9
48 49 4 10
49 50 4 11
50 51 4 12

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