Variance-covariance matrix with multiple columns - python-3.x

I have the following data:
at_score atp_1 atp_2 atp_3 g_date g_id g_time ht_diff ht_score htp_1 htp_2 htp_3
0 0 6 7 8 11/16/18 1 0 0 0 1 2 3
1 13 6 7 9 11/16/18 1 15 2 15 1 2 3
2 20 7 8 10 11/16/18 1 18 2 22 3 4 5
3 40 7 8 6 11/16/18 1 33 5 45 4 1 2
4 65 8 7 6 11/16/18 1 60 -3 62 1 2 3
5 0 6 7 8 11/20/18 2 0 0 0 1 2 3
6 10 9 7 8 11/20/18 2 7 -4 6 4 2 3
7 26 6 10 7 11/20/18 2 24 -1 25 1 5 4
8 40 9 7 8 11/20/18 2 42 5 45 1 2 5
9 65 6 7 10 11/20/18 2 60 5 70 1 5 2
where at_score, ht_score are the away & home team's score on a particular date (g_date), in a particular game (g_id), & at a particular time in the game (g_time). ht_diff represents the home team's score differential (ht_score - at_score). Finally, and for my purposes most importantly, atp_1, atp_2, atp_3 are the 3 away players who are playing at that point. htp_1, htp_2, htp_3 are their home team counterparts.
What I'd like to calculate is the variance-covariance matrix for each of the home & away team players based on how the ht_diff, ht_score & at_score changed while they were playing and the players they were playing with. For example away player 6 played with players 7 & 8 for the first 13 minutes of g_id 1 (ht_diff = 2 for this period) & the last 27 minutes (ht_diff = -3).
In the end I have about 2.5 million observations (as well as 10 players playing at a time) so finding a 'easy' to calculate this would be extremely helpful.

Related

How to get rows when specific column value is continous for certain number of rows

I want to extract rows when the column x value remains the same for more than five consecutive rows.
x x2
0 5 5
1 4 5
2 10 6
3 10 5
4 10 6
5 10 78
6 10 89
7 10 78
8 10 98
9 10 8
10 10 56
11 60 45
12 10 65
Desired_output:
x x2
0 10 6
1 10 5
2 10 6
3 10 78
4 10 89
5 10 78
6 10 98
7 10 8
8 10 56

You can use .shift + .cumsum to identify the blocks of consecutive rows where column x value remains same, then group the dataframe on these blocks and transform using count to identify the groups which have greater than 5 consecutive same values in x:
b = df['x'].ne(df['x'].shift()).cumsum()
df_out = df[df['x'].groupby(b).transform('count').gt(5)]
Details:
>>> b
0 1
1 2
2 3
3 3
4 3
5 3
6 3
7 3
8 3
9 3
10 3
11 4
12 5
Name: x, dtype: int64
>>> df_out
x x2
2 10 6
3 10 5
4 10 6
5 10 78
6 10 89
7 10 78
8 10 98
9 10 8
10 10 56

you can use shift to compare the next row and then take cumulative sum to compare if the repeat is greater than 5, then group on x and transform any then mask with the condition to unselect rows where condition does not match.
c = df['x'].eq(df['x'].shift())
out = df[c.cumsum().gt(5).groupby(df['x']).transform('any') & (c|c.shift(-1))]
print(out)
x x2
2 10 6
3 10 5
4 10 6
5 10 78
6 10 89
7 10 78
8 10 98
9 10 8
10 10 56

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 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

How to find the numver of duplicate lines, each line contains a few numbers seperated by spaces

Suppose i have a file like this...
4 2 8 2 12 3 18 2 22 2 26 2 28 3 30 2
4 3 10 2 14 2 18 2 20 3 22 2 28 2 32 2
2 3 10 3 12 2 16 2 18 3 20 2 24 2 26 3
1 3 3 3 17 3 19 3 26 2 28 2 30 2 32 2
4 2 8 2 12 3 18 2 22 2 26 2 28 3 30 2
the first and the last line are the same in the input...
I want the output to be like ...
4 2 8 2 12 3 18 2 22 2 26 2 28 3 30 2 2
4 3 10 2 14 2 18 2 20 3 22 2 28 2 32 2 1
2 3 10 3 12 2 16 2 18 3 20 2 24 2 26 3 1
1 3 3 3 17 3 19 3 26 2 28 2 30 2 32 2 1
The extra last coloum in the output simply specifies the extra number of lines.....
how can i do this in bash...
i know the sort command but it only works with one number per line....

Coming from sehe's suggestion, what about this?
sort your_file | uniq -c | awk '{for(i=2;i<=NF;i++) printf $i"\t"; printf $1"\n"}'
Output:
1 3 3 3 17 3 19 3 26 2 28 2 30 2 32 2 1
2 3 10 3 12 2 16 2 18 3 20 2 24 2 26 3 1
4 2 8 2 12 3 18 2 22 2 26 2 28 3 30 2 2
4 3 10 2 14 2 18 2 20 3 22 2 28 2 32 2 1

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