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Pandas Multilevel header, creating multiple rows for entries within a line

I have a multilevel column pandas dataframe with orders from an online retailer. Each line has info from a single order. This results in multiple items within the same row. I need to create a single row for each item sold in the orders. Because of this there are hundreds of columns.
orderID orderinfo orderline orderline orderline orderline
nan orderaspect1 0 0 1 1
nan nan itemaspect1 itemaspect2 itemaspect1 itemaspect2
0 1 2 3 4 5 6
1 10 20 30 40 50 60
2 100 200 300 400 500 600
each row with the same number under orderline needs to have its own row which includes the info from order iD and all the order aspects. It needs to look something like this.
orderID orderinfo orderline orderline item #
nan nan itemaspect1 itemaspect2
0 1 2 3 4 0
1 10 20 30 40 0
2 100 200 300 400 0
3 1 2 5 6 1
4 10 20 50 60 1
5 100 200 500 600 1
This way there is a row for each ITEM instead of a row for each ORDER.
I've tried using melt and stack to unpivot followed by pivoting but I've run into issue with that and the multiindex columns. Nothing is formatting it correctly.
[edit]
It should look like this using code.
i = pd.DatetimeIndex(['2011-03-31', '2011-04-01', '2011-04-04', '2011-04-05',
'2011-04-06', '2011-04-07', '2011-04-08', '2011-04-11',
'2011-04-12', '2011-04-13'])
cols = pd.MultiIndex.from_product([['orderID', "orderlines"],['order data', 0, 1]])
df = pd.DataFrame(np.random.randint(10, size=(len(i), 6)),index=i, columns=cols)

Trying to group by, then sort a dataframe based on multiple values [duplicate]

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])

Pandas : Mapping one column values using other dataframe column

I have two dataframes as described above
I would like to create in the second table an additional feature (Col_to_create) related to the value of feature A.
Table 2 has more than 800 000 samples so that I ask for a faster way to do that.
First table:
a b
1 100
2 400
3 500
Second table:
id Refer_to_A Col_to_create
0 3 500
1 1 100
2 3 500
3 2 400
4 1 100

You can use the method map:
df2['Col_to_create'] = df2['Refer_to_A'].map(df1.set_index('a')['b'])
Output:
Refer_to_A Col_to_create
id
0 3 500
1 1 100
2 3 500
3 2 400
4 1 100

One possible way is you can apply the function on new column of the dataset :
If your dataset is :
dataframe_a = pd.DataFrame({'a': [1,2,3], 'b': [100,400,500]})
dataframe_b = pd.DataFrame({'Refer_to_A': [3,1,3,2,1]})
You can try something like :
dataframe_b['Col_to_create'] = dataframe_b['Refer_to_A'].apply(lambda col: dataframe_a['b'][col-1])
output:
Refer_to_A Col_to_create
0 3 500
1 1 100
2 3 500
3 2 400
4 1 100

Compare corresponding columns with each other and store the result in a new column

I had a data which I pivoted using pivot table method , now the data looks like this:
rule_id a b c
50211 8 0 0
50249 16 0 3
50378 0 2 0
50402 12 9 6
I have set 'rule_id' as index. Now I compared one column to it's corresponding column and created another column with it's result. The idea is if the first column has a value other than 0 and the second column , to which the first column is compared to ,has 0 , then 100 should be updated in the newly created column, but if the situation is vice-versa then 'Null' should be updated. If both column have 0 , then also 'Null' should be updated. If the last column has value 0 , then 'Null' should be updated and other than 0 , then 100 should be updated. But if both the columns have values other than 0(like in the last row of my data) , then the comparison should be like this for column a and b:
value_of_b/value_of_a *50 + 50
and for column b and c:
value_of_c/value_of_b *25 + 25
and similarly if there are more columns ,then the multiplication and addition value should be 12.5 and so on.
I was able to achieve all the above things apart from the last result which is the division and multiplication stuff. I used this code:
m = df.eq(df.shift(-1, axis=1))
arr = np.select([df ==0, m], [np.nan, df], 1*100)
df2 = pd.DataFrame(arr, index=df.index).rename(columns=lambda x: f'comp{x+1}')
df3 = df.join(df2)
df is the dataframe which stores my pivoted table data which I mentioned at the start. After using this code my data looks like this:
rule_id a b c comp1 comp2 comp3
50211 8 0 0 100 NaN NaN
50249 16 0 3 100 NaN 100
50378 0 2 0 NaN 100 NaN
50402 12 9 6 100 100 100
But I want the data to look like this:
rule_id a b c comp1 comp2 comp3
50211 8 0 0 100 NaN NaN
50249 16 0 3 100 NaN 100
50378 0 2 0 NaN 100 NaN
50402 12 9 6 87.5 41.67 100
If you guys can help me get the desired data , I would greatly appreciate it.
Edit:
This is how my data looks:

The problem is that the coefficient to use when building the new compx column does not depend only on the columns position. In fact in each row it is reset to its maximum of 50 after each 0 value and is half of previous one after a non 0 value. Those resetable series are hard to vectorize in pandas, especially in rows. Here I would build a companion dataframe holding only those coefficients, and use directly the numpy underlying arrays to compute them as efficiently as possible. Code could be:
# transpose the dataframe to process columns instead of rows
coeff = df.T
# compute the coefficients
for name, s in coeff.items():
top = 100 # start at 100
r = []
for i, v in enumerate(s):
if v == 0: # reset to 100 on a 0 value
top=100
else:
top = top/2 # else half the previous value
r.append(top)
coeff.loc[:, name] = r # set the whole column in one operation
# transpose back to have a companion dataframe for df
coeff = coeff.T
# build a new column from 2 consecutive ones, using the coeff dataframe
def build_comp(col1, col2, i):
df['comp{}'.format(i)] = np.where(df[col1] == 0, np.nan,
np.where(df[col2] == 0, 100,
df[col2]/df[col1]*coeff[col1]
+coeff[col1]))
old = df.columns[0] # store name of first column
# Ok, enumerate all the columns (except first one)
for i, col in enumerate(df.columns[1:], 1):
build_comp(old, col, i)
old = col # keep current column name for next iteration
# special processing for last comp column
df['comp{}'.format(i+1)] = np.where(df[col] == 0, np.nan, 100)
With this initial dataframe:
date 2019-04-25 15:08:23 2019-04-25 16:14:14 2019-04-25 16:29:05 2019-04-25 16:36:32
rule_id
50402 0 0 9 0
51121 0 1 0 0
51147 0 1 0 0
51183 2 0 0 0
51283 0 12 9 6
51684 0 1 0 0
52035 0 4 3 2
it gives as expected:
date 2019-04-25 15:08:23 2019-04-25 16:14:14 2019-04-25 16:29:05 2019-04-25 16:36:32 comp1 comp2 comp3 comp4
rule_id
50402 0 0 9 0 NaN NaN 100.000000 NaN
51121 0 1 0 0 NaN 100.0 NaN NaN
51147 0 1 0 0 NaN 100.0 NaN NaN
51183 2 0 0 0 100.0 NaN NaN NaN
51283 0 12 9 6 NaN 87.5 41.666667 100.0
51684 0 1 0 0 NaN 100.0 NaN NaN
52035 0 4 3 2 NaN 87.5 41.666667 100.0

Ok, I think you can iterate over your dataframe df and use some if-else to get the desired output.
for i in range(len(df.index)):
if df.iloc[i,1]!=0 and df.iloc[i,2]==0: # column start from index 0
df.loc[i,'colname'] = 'whatever you want' # so rule_id is column 0
elif:
.
.
.

Python pandas: Append rows of DataFrame and delete the appended rows

import pandas as pd
df = pd.DataFrame({
'id':[1,2,3,4,5,6,7,8,9,10,11],
'text': ['abc','zxc','qwe','asf','efe','ert','poi','wer','eer','poy','wqr']})
I have a DataFrame with columns:
id text
1 abc
2 zxc
3 qwe
4 asf
5 efe
6 ert
7 poi
8 wer
9 eer
10 poy
11 wqr
I have a list L = [1,3,6,10] which contains list of id's.
I am trying to append the text column using a list such that, from my list first taking 1 and 3(first two values in a list) and appending text column in my DataFrame with id = 1 which has id's 2, then deleting rows with id column 2 similarly then taking 3 and 6 and then appending text column where id = 4,5 to id 3 and then delete rows with id = 4 and 5 and iteratively for elements in list (x, x+1)
My final output would look like this:
id text
1 abczxc # joining id 1 and 2
3 qweasfefe # joining id 3,4 and 5
6 ertpoiwereer # joining id 6,7,8,9
10 poywqr # joining id 10 and 11

You can use isin with cumsum for Series, which is use for groupby with apply join function:
s = df.id.where(df.id.isin(L)).ffill().astype(int)
df1 = df.groupby(s)['text'].apply(''.join).reset_index()
print (df1)
id text
0 1 abczxc
1 3 qweasfefe
2 6 ertpoiwereer
3 10 poywqr
It working because:
s = df.id.where(df.id.isin(L)).ffill().astype(int)
print (s)
0 1
1 1
2 3
3 3
4 3
5 6
6 6
7 6
8 6
9 10
10 10
Name: id, dtype: int32

I changed the values not in list to np.nan and then ffill and groupby. Though #Jezrael's approach is much better. I need to remember to use cumsum:)
l = [1,3,6,10]
df.id[~df.id.isin(l)] = np.nan
df = df.ffill().groupby('id').sum()
text
id
1.0 abczxc
3.0 qweasfefe
6.0 ertpoiwereer
10.0 poywqr

Use pd.cut to create you bins then groupby with a lambda function to join your text in that group.
df.groupby(pd.cut(df.id,L+[np.inf],right=False, labels=[i for i in L])).apply(lambda x: ''.join(x.text))
EDIT:
(df.groupby(pd.cut(df.id,L+[np.inf],
right=False,
labels=[i for i in L]))
.apply(lambda x: ''.join(x.text)).reset_index().rename(columns={0:'text'}))
Output:
id text
0 1 abczxc
1 3 qweasfefe
2 6 ertpoiwereer
3 10 poywqr