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How to identify one column with continuous number and same value of another column?

I have a DataFrame with two columns A and B.
I want to create a new column named C to identify the continuous A with the same B value.
Here's an example
import pandas as pd
df = pd.DataFrame({'A':[1,2,3,5,6,10,11,12,13,18], 'B':[1,1,2,2,3,3,3,3,4,4]})
I found a similar question, but that method only identifies the continuous A regardless of B.
df['C'] = df['A'].diff().ne(1).cumsum().sub(1)
I have tried to groupby B and apply the function like this:
df['C'] = df.groupby('B').apply(lambda x: x['A'].diff().ne(1).cumsum().sub(1))
However, it doesn't work: TypeError: incompatible index of inserted column with frame index.
The expected output is
A B C
1 1 0
2 1 0
3 2 1
5 2 2
6 3 3
10 3 4
11 3 4
12 3 4
13 4 5
18 4 6

Let's create a sequential counter using groupby, diff and cumsum then factorize to reencode the counter
df['C'] = df.groupby('B')['A'].diff().ne(1).cumsum().factorize()[0]
Result
A B C
0 1 1 0
1 2 1 0
2 3 2 1
3 5 2 2
4 6 3 3
5 10 3 4
6 11 3 4
7 12 3 4
8 13 4 5
9 18 4 6

Use DataFrameGroupBy.diff with compare not equal 1 and Series.cumsum, last subtract 1:
df['C'] = df.groupby('B')['A'].diff().ne(1).cumsum().sub(1)
print (df)
A B C
0 1 1 0
1 2 1 0
2 3 2 1
3 5 2 2
4 6 3 3
5 10 3 4
6 11 3 4
7 12 3 4
8 13 4 5
9 18 4 6

Sort a subset of columns of a pandas dataframe alphabetically by column name

I'm having trouble finding the solution to a fairly simple problem.
I would like to alphabetically arrange certain columns of a pandas dataframe that has over 100 columns (i.e. so many that I don't want to list them manually).
Example df:
import pandas as pd
subject = [1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,4,4,4,4,4,4]
timepoint = [1,2,3,4,5,6,1,2,3,4,5,6,1,2,4,1,2,3,4,5,6]
c = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
d = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
a = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
b = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
df = pd.DataFrame({'subject':subject,
'timepoint':timepoint,
'c':c,
'd':d,
'a':a,
'b':b})
df.head()
subject timepoint c d a b
0 1 1 2 2 2 2
1 1 2 3 3 3 3
2 1 3 4 4 4 4
3 1 4 5 5 5 5
4 1 5 6 6 6 6
How could I rearrange the column names to generate a df.head() that looks like this:
subject timepoint a b c d
0 1 1 2 2 2 2
1 1 2 3 3 3 3
2 1 3 4 4 4 4
3 1 4 5 5 5 5
4 1 5 6 6 6 6
i.e. keep the first two columns where they are and then alphabetically arrange the remaining column names.
Thanks in advance.

You can split your your dataframe based on column names, using normal indexing operator [], sort alphabetically the other columns using sort_index(axis=1), and concat back together:
>>> pd.concat([df[['subject','timepoint']],
df[df.columns.difference(['subject', 'timepoint'])]\
.sort_index(axis=1)],ignore_index=False,axis=1)
subject timepoint a b c d
0 1 1 2 2 2 2
1 1 2 3 3 3 3
2 1 3 4 4 4 4
3 1 4 5 5 5 5
4 1 5 6 6 6 6
5 1 6 7 7 7 7
6 2 1 3 3 3 3
7 2 2 4 4 4 4
8 2 3 1 1 1 1
9 2 4 2 2 2 2
10 2 5 3 3 3 3
11 2 6 4 4 4 4
12 3 1 5 5 5 5
13 3 2 4 4 4 4
14 3 4 5 5 5 5
15 4 1 8 8 8 8
16 4 2 4 4 4 4
17 4 3 5 5 5 5
18 4 4 6 6 6 6
19 4 5 2 2 2 2
20 4 6 3 3 3 3

Specify the first two columns you want to keep (or determine them from the data), then sort all of the other columns. Use .loc with the correct list to then "sort" the DataFrame.
import numpy as np
first_cols = ['subject', 'timepoint']
#first_cols = df.columns[0:2].tolist() # OR determine first two
other_cols = np.sort(df.columns.difference(first_cols)).tolist()
df = df.loc[:, first_cols+other_cols]
print(df.head())
subject timepoint a b c d
0 1 1 2 2 2 2
1 1 2 3 3 3 3
2 1 3 4 4 4 4
3 1 4 5 5 5 5
4 1 5 6 6 6 6

You can try getting the dataframe columns as a list, rearrange them, and assign it back to the dataframe using df = df[cols]
import pandas as pd
subject = [1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,4,4,4,4,4,4]
timepoint = [1,2,3,4,5,6,1,2,3,4,5,6,1,2,4,1,2,3,4,5,6]
c = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
d = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
a = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
b = [2,3,4,5,6,7,3,4,1,2,3,4,5,4,5,8,4,5,6,2,3]
df = pd.DataFrame({'subject':subject,
'timepoint':timepoint,
'c':c,
'd':d,
'a':a,
'b':b})
cols = df.columns.tolist()
cols = cols[:2] + sorted(cols[2:])
df = df[cols]

Is it possible to obtain groupby style counts without collapsing Pandas DataFrame?

I have a DataFrame with 9 columns, and I'm trying to add a column of counts of unique values based on the first 3 columns (e.g. Cols A, B, and C, must match to count as a unique value , but the remaining columns can vary. I attempted to do this as with groupby:
df = pd.DataFrame(resultsFile500.groupby(['chr','start','end']).size().reset_index().rename(columns={0:'count'}))
This returns a DataFrame with 5 columns, and the counts are what I want. However, I also need values from the original data frame, so what I have been trying to do is somehow get those values of counts as a column in the original df. So, this would mean that if two rows in columns chr, start, and end, had identical values, the counts column would be 2 in both rows, but they would not be collapsed to one row. Is there an easy solution here that I'm missing, or do I need to hack something together?

You can use .transform to get non-collapsing behavior:
>>> df
a b c d e
0 3 4 1 3 0
1 3 1 4 3 0
2 4 3 3 2 1
3 3 4 1 4 0
4 0 4 3 3 2
5 1 2 0 4 1
6 3 1 4 2 1
7 0 4 3 4 0
8 1 3 0 1 1
9 3 4 1 2 1
>>> df.groupby(['a','b','c']).transform('count')
d e
0 3 3
1 2 2
2 1 1
3 3 3
4 2 2
5 1 1
6 2 2
7 2 2
8 1 1
9 3 3
>>>
Note, i'll have to choose an arbitrary column from the .transform result, but then just do:
>>> df['unique_count'] = df.groupby(['a','b','c']).transform('count')['d']
>>> df
a b c d e unique_count
0 3 4 1 3 0 3
1 3 1 4 3 0 2
2 4 3 3 2 1 1
3 3 4 1 4 0 3
4 0 4 3 3 2 2
5 1 2 0 4 1 1
6 3 1 4 2 1 2
7 0 4 3 4 0 2
8 1 3 0 1 1 1
9 3 4 1 2 1 3

Pandas reverse column values groupwise

I want to reverse a column values in my dataframe, but only on a individual "groupby" level. Below you can find a minimal demonstration example, where I want to "flip" values that belong the same letter A,B or C:
df = pd.DataFrame({"group":["A","A","A","B","B","B","B","C","C"],
"value": [1,3,2,4,4,2,3,2,5]})
group value
0 A 1
1 A 3
2 A 2
3 B 4
4 B 4
5 B 2
6 B 3
7 C 2
8 C 5
My desired output looks like this: (column is added instead of replaced only for the brevity purposes)
group value value_desired
0 A 1 2
1 A 3 3
2 A 2 1
3 B 4 3
4 B 4 2
5 B 2 4
6 B 3 4
7 C 2 5
8 C 5 2
As always, when I don't see a proper vector-style approach, I end messing with loops just for the sake of final output, but my current code hurts me very much:
for i in list(set(df["group"].values.tolist())):
reversed_group = df.loc[df["group"]==i,"value"].values.tolist()[::-1]
df.loc[df["group"]==i,"value_desired"] = reversed_group
Pandas gurus, please show me the way :)

You can use transform
In [900]: df.groupby('group')['value'].transform(lambda x: x[::-1])
Out[900]:
0 2
1 3
2 1
3 3
4 2
5 4
6 4
7 5
8 2
Name: value, dtype: int64
Details
In [901]: df['value_desired'] = df.groupby('group')['value'].transform(lambda x: x[::-1])
In [902]: df
Out[902]:
group value value_desired
0 A 1 2
1 A 3 3
2 A 2 1
3 B 4 3
4 B 4 2
5 B 2 4
6 B 3 4
7 C 2 5
8 C 5 2

Index matching with multiple columns in python

I have two pandas dataframe with different size. two dataframe looks like
df1 =
x y data
1 2 5
2 2 7
5 3 9
3 5 2
and another dataframe looks like:
df2 =
x y value
5 3 7
1 2 4
3 5 2
7 1 4
4 6 5
2 2 1
7 5 8
I am trying to merge these two dataframe so that the final dataframe expected to have same combination of x and y with respective value. I am expecting final dataframe in this format:
x y data value
1 2 5 4
2 2 7 1
5 3 9 7
3 5 2 2
I tride this code but not getting expected results.
dfB.set_index('x').loc[dfA.x].reset_index()

Use merge, by default how='inner' so it can be omit and if join only on same columns parameter on can be omit too:
print (pd.merge(df1,df2))
x y data value
0 1 2 5 4
1 2 2 7 1
2 5 3 9 7
3 3 5 2 2
If in real data are multiple same column names use:
print (pd.merge(df1,df2, on=['x','y']))
x y data value
0 1 2 5 4
1 2 2 7 1
2 5 3 9 7
3 3 5 2 2

df1.merge(df2,by='x')
This will do