Following are the dataframes I have
score_df
col1_id col2_id score
1 2 10
5 6 20
records_df
date col_id
D1 6
D2 4
D3 1
D4 2
D5 5
D6 7
I would like to compute a score based on the following criteria:
When 2 occurs after 1 the score should be assigned 10 or when 1 occurs after 2, score should be assigned 10.
i.e when (1,2) gives a score 10 .. (2,1) also get the same score 10.
considering (1,2) . When 1 occurs first time we dont assign a score. We flag the row and wait for 2 to occur. When 2 occurs in the column we give the score 10.
considering (2,1). When 2 comes first. We assign value 0 and wait for 1 to occur. When 1 occurs, we give the score 10.
So, for the first time - dont assign the score and wait for the corresponding event to occur and then assign the score
So, my result dataframe should look something like this
result
date col_id score
D1 6 0 -- Eventhough 6 is there in score list, it occured for first time. So 0
D2 4 0 -- 4 is not even there in list
D3 1 0 -- 1 occurred for first time . So 0
D4 2 10 -- 1 occurred previously. 2 occurred now.. we can assign 10.
D5 5 20 -- 6 occurred previously. we can assign 20
D6 7 0 -- 7 is not in the list
I have around 100k rows in both score_df and record_df. Looping and assigning score is taking the time. Can someone help with logic without looping the entire dataframe?
From what i understand , you can try melt for unpivotting and then merge. keeping the index from the melted df , we check where the index is duplicated , and then return score from the merge else 0.
m = score_df.reset_index().melt(['index','uid','score'],
var_name='col_name',value_name='col_id')
final = records_df.merge(m.drop('col_name',1),on=['uid','col_id'],how='left')
c = final.duplicated(['index']) & final['index'].notna()
final = final.drop('index',1).assign(score=lambda x: x['score'].where(c,0))
print(final)
uid date col_id score
0 123 D1 6 0.0
1 123 D2 4 0.0
2 123 D3 1 0.0
3 123 D4 2 10.0
4 123 D5 5 20.0
5 123 D6 7 0.0
Related
Suppose we have a DataFrame like this, only with many, many more index A values:
df = pd.DataFrame([[1,2,1,2],
[1,1,2,2],
[2,2,1,0],
[1,2,1,2],
[2,1,1,2] ], columns=['A','B','c1','c2'])
df.groupby(['A','B']).sum()
## result
c1 c2
A B
1 1 2 2
2 2 4
2 1 1 2
2 1 0
How can I get a data frame that consists of the difference between rows, by the second level of the index, level B?
The output here would be
A c1 c2
1 0 -2
2 0 2
Note In my particular use case, I have a lot of column A values, so I can write out the value for A explicitly.
Check diff and dropna
g = df.groupby(['A','B'])[['c1','c2']].sum()
g = g.groupby(level=0).diff().dropna()
g
Out[25]:
c1 c2
A B
1 2 0.0 2.0
2 2 0.0 -2.0
Assigning the first grouping to result variable:
result = df.groupby(['A','B']).sum()
You could use a pipe operation with nth:
result.groupby('A').pipe(lambda df: df.nth(0) - df.nth(-1))
c1 c2
A
1 0 -2
2 0 2
A simpler option, in my opinion, would be to use agg combined with numpy's ufunc reduce, as this covers scenarios where you have more than two rows:
result.groupby('A').agg(np.subtract.reduce)
c1 c2
A
1 0 -2
2 0 2
Having df of probabilities distribution, I get max probability for rows with df.idxmax(axis=1) like this:
df['1k-th'] = df.idxmax(axis=1)
and get the following result:
(scroll the tables to the right if you can not see all the columns)
0 1 2 3 4 5 6 1k-th
0 0.114869 0.020708 0.025587 0.028741 0.031257 0.031619 0.747219 6
1 0.020206 0.012710 0.010341 0.012196 0.812495 0.113863 0.018190 4
2 0.023585 0.735475 0.091795 0.021683 0.027581 0.054217 0.045664 1
3 0.009834 0.009175 0.013165 0.016014 0.015507 0.899115 0.037190 5
4 0.023357 0.736059 0.088721 0.021626 0.027341 0.056289 0.046607 1
the question is how to get the 2-th, 3th, etc probabilities, so that I get the following result?:
0 1 2 3 4 5 6 1k-th 2-th
0 0.114869 0.020708 0.025587 0.028741 0.031257 0.031619 0.747219 6 0
1 0.020206 0.012710 0.010341 0.012196 0.812495 0.113863 0.018190 4 3
2 0.023585 0.735475 0.091795 0.021683 0.027581 0.054217 0.045664 1 4
3 0.009834 0.009175 0.013165 0.016014 0.015507 0.899115 0.037190 5 4
4 0.023357 0.736059 0.088721 0.021626 0.027341 0.056289 0.046607 1 2
Thank you!
My own solution is not the prettiest, but does it's job and works fast:
for i in range(7):
p[f'{i}k'] = p[[0,1,2,3,4,5,6]].idxmax(axis=1)
p[f'{i}k_v'] = p[[0,1,2,3,4,5,6]].max(axis=1)
for x in range(7):
p[x] = np.where(p[x]==p[f'{i}k_v'], np.nan, p[x])
The loop does:
finds the largest value and it's column index
drops the found value (sets to nan)
again
finds the 2nd largest value
drops the found value
etc ...
in my data frame I want to iterrows() of two columns but want to save result in 1 column.for example df is
x y
5 10
30 445
70 32
expected output is
points sequence
5 1
10 2
30 1
445 2
I know about iterrows() but it saved out put in two different columns.How can I get expected output and is there any way to generate sequence number according to condition? any help will be appreciated.
First never use iterrows, because really slow.
If want 1, 2 sequence by number of columns convert values to numy array by DataFrame.to_numpy and add numpy.ravel, then for sequence use numpy.tile:
df = pd.DataFrame({'points': df.to_numpy().ravel(),
'sequence': np.tile([1,2], len(df))})
print (df)
points sequence
0 5 1
1 10 2
2 30 1
3 445 2
4 70 1
5 32 2
Do this way:
>>> pd.DataFrame([i[1] for i in df.iterrows()])
points sequence
0 5 1
1 10 2
2 30 1
3 445 2
Suppose I have pandas DataFrame like this:
df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4], 'value':[1,2,3,1,2,3,4,1,1]})
which looks like:
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
I want to get a new DataFrame with top 2 records for each id, like this:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
I can do it with numbering records within group after groupby:
dfN = df.groupby('id').apply(lambda x:x['value'].reset_index()).reset_index()
which looks like:
id level_1 index value
0 1 0 0 1
1 1 1 1 2
2 1 2 2 3
3 2 0 3 1
4 2 1 4 2
5 2 2 5 3
6 2 3 6 4
7 3 0 7 1
8 4 0 8 1
then for the desired output:
dfN[dfN['level_1'] <= 1][['id', 'value']]
Output:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
But is there more effective/elegant approach to do this? And also is there more elegant approach to number records within each group (like SQL window function row_number()).
Did you try
df.groupby('id').head(2)
Output generated:
id value
id
1 0 1 1
1 1 2
2 3 2 1
4 2 2
3 7 3 1
4 8 4 1
(Keep in mind that you might need to order/sort before, depending on your data)
EDIT: As mentioned by the questioner, use
df.groupby('id').head(2).reset_index(drop=True)
to remove the MultiIndex and flatten the results:
id value
0 1 1
1 1 2
2 2 1
3 2 2
4 3 1
5 4 1
Since 0.14.1, you can now do nlargest and nsmallest on a groupby object:
In [23]: df.groupby('id')['value'].nlargest(2)
Out[23]:
id
1 2 3
1 2
2 6 4
5 3
3 7 1
4 8 1
dtype: int64
There's a slight weirdness that you get the original index in there as well, but this might be really useful depending on what your original index was.
If you're not interested in it, you can do .reset_index(level=1, drop=True) to get rid of it altogether.
(Note: From 0.17.1 you'll be able to do this on a DataFrameGroupBy too but for now it only works with Series and SeriesGroupBy.)
Sometimes sorting the whole data ahead is very time consuming.
We can groupby first and doing topk for each group:
g = df.groupby(['id']).apply(lambda x: x.nlargest(topk,['value'])).reset_index(drop=True)
df.groupby('id').apply(lambda x : x.sort_values(by = 'value', ascending = False).head(2).reset_index(drop = True))
Here sort values ascending false gives similar to nlargest and True gives similar to nsmallest.
The value inside the head is the same as the value we give inside nlargest to get the number of values to display for each group.
reset_index is optional and not necessary.
This works for duplicated values
If you have duplicated values in top-n values, and want only unique values, you can do like this:
import pandas as pd
ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])
id first_name last_name department salary
24 12 Shandler Bing Audit 110000
25 14 Jason Tom Audit 100000
26 16 Celine Anston Audit 100000
27 15 Michale Jackson Audit 70000
If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:
(df.groupby('department')['salary']
.apply(lambda ser: ser.drop_duplicates().nlargest(3))
.droplevel(level=1)
.sort_index()
.reset_index()
)
This gives
department salary
0 Audit 110000
1 Audit 100000
2 Audit 70000
3 Management 250000
4 Management 200000
5 Management 150000
6 Sales 220000
7 Sales 200000
8 Sales 150000
To get the first N rows of each group, another way is via groupby().nth[:N]. The outcome of this call is the same as groupby().head(N). For example, for the top-2 rows for each id, call:
N = 2
df1 = df.groupby('id', as_index=False).nth[:N]
To get the largest N values of each group, I suggest two approaches.
First sort by "id" and "value" (make sure to sort "id" in ascending order and "value" in descending order by using the ascending parameter appropriately) and then call groupby().nth[].
N = 2
df1 = df.sort_values(by=['id', 'value'], ascending=[True, False])
df1 = df1.groupby('id', as_index=False).nth[:N]
Another approach is to rank the values of each group and filter using these ranks.
# for the entire rows
N = 2
msk = df.groupby('id')['value'].rank(method='first', ascending=False) <= N
df1 = df[msk]
# for specific column rows
df1 = df.loc[msk, 'value']
Both of these are much faster than groupby().apply() and groupby().nlargest() calls as suggested in the other answers on here(1, 2, 3). On a sample with 100k rows and 8000 groups, a %timeit test showed that it was 24-150 times faster than those solutions.
Also, instead of slicing, you can also pass a list/tuple/range to a .nth() call:
df.groupby('id', as_index=False).nth([0,1])
# doesn't even have to be consecutive
# the following returns 1st and 3rd row of each id
df.groupby('id', as_index=False).nth([0,2])
Below is some data:
Test Day1 Day2 Score
A 1 2 100
B 1 3 62
C 3 4 90
D 2 4 20
E 4 5 80
I am trying to take the values from column 'day' and 'day2' and use them to select the row number for the column score. For example for Test A I would like to find the sum of 100 and 62 because that is the values of the first and second rows of score. Test B I would like to find the sum of 100, 62 and 90.
Does anyone have any ideas on how to go about doing this? I am looking to use something similar to the indirect function in Excel? Thank You
The trick is to convert variable "Score" as a row. Could not think of an easy way how to avoid SAVE/GET - room for improvements.
file handle tmp
/name = "C:\DATA\Temp".
***.
data list free /Test (a1) Day1 (f8) Day2 (f8) Score (f8).
begin data
A 1 2 100
B 1 3 62
C 3 4 90
D 2 4 20
E 4 5 80
end data.
comp f = 1.
var wid all (12).
save out "tmp\data.sav".
***.
get "tmp\data.sav"
/keep score.
flip.
comp f = 1.
match files
/file "tmp\data.sav"
/table *
/by f
/drop case_lbl.
comp stat = 0.
do rep var = var001 to var005
/k = 1 to 5.
if range(k, Day1, Day2) stat = sum(stat, var).
end rep.
list Test Day1 Day2 Score stat.
The result:
Test Day1 Day2 Score stat
A 1 2 100 162
B 1 3 62 252
C 3 4 90 110
D 2 4 20 172
E 4 5 80 100
Number of cases read: 5 Number of cases listed: 5