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create a new column using previous row of other column consider a duplicate as well as a based on group [duplicate]

I have a Pandas dataframe, and I want to create a new column whose values are that of another column, shifted down by one row. The last row should show NaN.
The catch is that I want to do this by group, with the last row of each group showing NaN. NOT have the last row of a group "steal" a value from a group that happens to be adjacent in the dataframe.
My attempted implementation is quite shamefully broken, so I'm clearly misunderstanding something fundamental.
df['B_shifted'] = df.groupby(['A'])['B'].transform(lambda x:x.values[1:])

Newer versions of pandas can now perform a shift on a group:
df['B_shifted'] = df.groupby(['A'])['B'].shift(1)
Note that when shifting down, it's the first row that has NaN.

Shift works on the output of the groupby clause:
>>> df = pandas.DataFrame(numpy.random.randint(1,3, (10,5)), columns=['a','b','c','d','e'])
>>> df
a b c d e
0 2 1 2 1 1
1 2 1 1 1 1
2 1 2 2 1 2
3 1 2 1 1 2
4 2 2 1 1 2
5 2 2 2 2 1
6 2 2 1 1 1
7 2 2 2 1 1
8 2 2 2 2 1
9 2 2 2 2 1
for k, v in df.groupby('a'):
print k
print 'normal'
print v
print 'shifted'
print v.shift(1)
1
normal
a b c d e
2 1 2 2 1 2
3 1 2 1 1 2
shifted
a b c d e
2 NaN NaN NaN NaN NaN
3 1 2 2 1 2
2
normal
a b c d e
0 2 1 2 1 1
1 2 1 1 1 1
4 2 2 1 1 2
5 2 2 2 2 1
6 2 2 1 1 1
7 2 2 2 1 1
8 2 2 2 2 1
9 2 2 2 2 1
shifted
a b c d e
0 NaN NaN NaN NaN NaN
1 2 1 2 1 1
4 2 1 1 1 1
5 2 2 1 1 2
6 2 2 2 2 1
7 2 2 1 1 1
8 2 2 2 1 1
9 2 2 2 2 1

#EdChum's comment is a better answer to this question, so I'm posting it here for posterity:
df['B_shifted'] = df.groupby(['A'])['B'].transform(lambda x:x.shift())
or similarly
df['B_shifted'] = df.groupby(['A'])['B'].transform('shift').
The former notation is more flexible, of course (e.g. if you want to shift by 2).

Creating a column that assigns max value of set of rows by condition to all rows in that group

I have a dataframe that looks like this:
data metadata
A 0
A 1
A 2
A 3
A 4
B 0
B 1
B 2
A 0
A 1
B 0
A 0
A 1
B 0
df.data contains two different categories, A and B. df.metadata stores a running count the number of times a category appears consecutively before the category changes. I want to create a column consecutive_count that assigns the max value of metadata per consecutive group to every row in that group. It should look like this:
data metadata consecutive_count
A 0 4
A 1 4
A 2 4
A 3 4
A 4 4
B 0 2
B 1 2
B 2 2
A 0 1
A 1 1
B 0 0
A 0 1
A 1 1
B 0 0
Please advise. Thank you.

Method 1:
You may try transform max on groupby of each group of data
s = df.data.ne(df.data.shift()).cumsum()
df['consecutive_count'] = df.groupby(s).metadata.transform('max')
Out[96]:
data metadata consecutive_count
0 A 0 4
1 A 1 4
2 A 2 4
3 A 3 4
4 A 4 4
5 B 0 2
6 B 1 2
7 B 2 2
8 A 0 1
9 A 1 1
10 B 0 0
11 A 0 1
12 A 1 1
13 B 0 0
Method 2:
Since metadata is sorted per group, you may reverse dataframe and do groupby cummax
s = df.data.ne(df.data.shift()).cumsum()
df['consecutive_count'] = df[::-1].groupby(s).metadata.cummax()
Out[101]:
data metadata consecutive_count
0 A 0 4
1 A 1 4
2 A 2 4
3 A 3 4
4 A 4 4
5 B 0 2
6 B 1 2
7 B 2 2
8 A 0 1
9 A 1 1
10 B 0 0
11 A 0 1
12 A 1 1
13 B 0 0

Python Counting Same Values For Specific Columns

If i have a dataframe;
A B C D
1 1 2 2 1
2 1 1 2 1
3 3 1 0 1
4 2 4 4 4
I want to make addition B and C columns and counting whether or not the same values with D columns. Desired output is;
A B C B+C D
1 1 2 2 4 1
2 1 1 2 3 1
3 3 1 0 1 1
4 2 4 4 8 4
There are 3 different values compare the "B+C" and "D".
Could you please help me about this?

You could do something like:
df.B.add(df.C).ne(df.D).sum()
# 3
If you need to add the column:
df['B+C'] = df.B.add(df.C)
diff = df['B+C'].ne(df.D).sum()
print(f'There are {diff} different values compare the "B+C" and "D"')
#There are 3 different values compare the "B+C" and "D"

df.insert(3,'B+C', df['B']+df['C'])
3 is the index
df.head()
A B C B+C D
0 1 2 2 4 1
1 1 1 2 3 1
2 3 1 0 1 1
3 2 4 4 8 4
After that you can follow the steps of #yatu
df['B+C'].ne(df['D'])
0 True
1 True
2 False
3 True dtype: bool
df['B+C'].ne(df['D']).sum()
3

Use groupby and merge to create new column in pandas

So I have a pandas dataframe that looks something like this.
name is_something
0 a 0
1 b 1
2 c 0
3 c 1
4 a 1
5 b 0
6 a 1
7 c 0
8 a 1
Is there a way to use groupby and merge to create a new column that gives the number of times a name appears with an is_something value of 1 in the whole dataframe? The updated dataframe would look like this:
name is_something no_of_times_is_something_is_1
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3
I know you can just loop through the dataframe to do this but I'm looking for a more efficient way because the dataset I'm working with is quite large. Thanks in advance!

If there are only 0 and 1 values in is_something column only use sum with GroupBy.transform for new column filled by aggregate values:
df['new'] = df.groupby('name')['is_something'].transform('sum')
print (df)
name is_something new
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3
If possible multiple values first compare by 1, convert to integer and then use transform with sum:
df['new'] = df['is_something'].eq(1).view('i1').groupby(df['name']).transform('sum')

Or we just map it
df['New']=df.name.map(df.query('is_something ==1').groupby('name')['is_something'].sum())
df
name is_something New
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3

You could do:
df['new'] = df.groupby('name')['is_something'].transform(lambda xs: xs.eq(1).sum())
print(df)
Output
name is_something new
0 a 0 3
1 b 1 1
2 c 0 1
3 c 1 1
4 a 1 3
5 b 0 1
6 a 1 3
7 c 0 1
8 a 1 3

cumulative number of unique elements for pandas dataframe

i have a pandas data frame
id tag
1 A
1 A
1 B
1 C
1 A
2 B
2 C
2 B
I want to add a column which computes the cumulative number of unique tags over at id level. More specifically, I would like to have
id tag count
1 A 1
1 A 1
1 B 2
1 C 3
1 A 3
2 B 1
2 C 2
2 B 2
For a given id, count will be non-decreasing. Thanks for your help!

I think this does what you want:
unique_count = df.drop_duplicates().groupby('id').cumcount() + 1
unique_count.reindex(df.index).ffill()
The +1 is because the count starts at zero. This only works if the dataframe is sorted by id. Was that intended? You can always sort beforehand.

You can find some other approaches in R and Python here
df = pd.DataFrame({'id':[1,1,1,1,1,2,2,2],'tag':["A","A", "B","C","A","B","C","B"]})
df['count']=df.groupby('id')['tag'].apply(lambda x: (~pd.Series(x).duplicated()).cumsum())
id tag count
0 1 A 1
1 1 A 1
2 1 B 2
3 1 C 3
4 1 A 3
5 2 B 1
6 2 C 2
7 2 B 2

How about this:
d['X'] = 1
d.groupby("Col").X.cumsum()

idt=[1,1,1,1,1,2,2,2]
tag=['A','A','B','C','A','B','C','B']
df=pd.DataFrame(tag,index=idt,columns=['tag'])
df=df.reset_index()
print(df)
index tag
0 1 A
1 1 A
2 1 B
3 1 C
4 1 A
5 2 B
6 2 C
7 2 B
df['uCnt']=df.groupby(['index','tag']).cumcount()+1
print(df)
index tag uCnt
0 1 A 1
1 1 A 2
2 1 B 1
3 1 C 1
4 1 A 3
5 2 B 1
6 2 C 1
7 2 B 2
df['uCnt']=df['uCnt']//df['uCnt']**2
print(df)
index tag uCnt
0 1 A 1
1 1 A 0
2 1 B 1
3 1 C 1
4 1 A 0
5 2 B 1
6 2 C 1
7 2 B 0
df['uCnt']=df.groupby(['index'])['uCnt'].cumsum()
print(df)
index tag uCnt
0 1 A 1
1 1 A 1
2 1 B 2
3 1 C 3
4 1 A 3
5 2 B 1
6 2 C 2
7 2 B 2
df=df.set_index('index')
print(df)
tag uCnt
index
1 A 1
1 A 1
1 B 2
1 C 3
1 A 3
2 B 1
2 C 2
2 B 2