Concatenate alternate scalar column to pandas based on condition - pandas

Question

Have a master dataframe and a tag list, as follows:
import pandas as pd
i = ['A'] * 2 + ['B'] * 3 + ['A'] * 4 + ['B'] * 5
master = pd.DataFrame(i, columns={'cat'})
tag = [0, 1]
How to insert a column of tags that is normal for cat: A, but reversed for cat: B? Expected output is:
cat tags
0 A 0
1 A 1
2 B 1
3 B 0
4 B 1
5 A 0
6 A 1
7 A 0
8 A 1
9 B 1
10 B 0
...

Answer 1

EDIT: Because is necessary processing each concsecutive group separately I try create general solution:
tag = ['a','b','c']
r = range(len(tag))
r1 = range(len(tag)-1, -1, -1)
print (dict(zip(r1, tag)))
{2: 'a', 1: 'b', 0: 'c'}
m1 = master['cat'].eq('A')
m2 = master['cat'].eq('B')
s = master['cat'].ne(master['cat'].shift()).cumsum()
master['tags'] = master.groupby(s).cumcount() % len(tag)
master.loc[m1, 'tags'] = master.loc[m1, 'tags'].map(dict(zip(r, tag)))
master.loc[m2, 'tags'] = master.loc[m2, 'tags'].map(dict(zip(r1, tag)))
print (master)
cat tags
0 A a
1 A b
2 B c
3 B b
4 B a
5 A a
6 A b
7 A c
8 A a
9 B c
10 B b
11 B a
12 B c
13 B b
Another approach is create DataFrame from tags and merge with left join:
tag = ['a','b','c']
s = master['cat'].ne(master['cat'].shift()).cumsum()
master['g'] = master.groupby(s).cumcount() % len(tag)
d = {'A': tag, 'B':tag[::-1]}
df = pd.DataFrame([(k,i,x)
for k, v in d.items()
for i, x in enumerate(v)], columns=['cat','g','tags'])
print (df)
cat g tags
0 A 0 a
1 A 1 b
2 A 2 c
3 B 0 c
4 B 1 b
5 B 2 a
master = master.merge(df, on=['cat','g'], how='left').drop('g', axis=1)
print (master)
cat tags
0 A a
1 A b
2 B c
3 B b
4 B a
5 A a
6 A b
7 A c
8 A a
9 B c
10 B b
11 B a
12 B c
13 B b
Idea is use numpy.tile for repeat tag values by number of matched values with integer division and then filtering by indexing and assign by both masks:
le = len(tag)
m1 = master['cat'].eq('A')
m2 = master['cat'].eq('B')
s1 = m1.sum()
s2 = m2.sum()
master.loc[m1, 'tags'] = np.tile(tag, s1 // le + le)[:s1]
#swapped order for m2 mask
master.loc[m2, 'tags'] = np.tile(tag[::-1], s2// le + le)[:s2]
print (master)
cat tags
0 A 0.0
1 A 1.0
2 B 1.0
3 B 0.0
4 B 1.0
5 A 0.0
6 A 1.0
7 A 0.0
8 A 1.0

Answer 2

IIUC, GroupBy.cumcount + Series.mod.
Then we invert the sequence where cat is B with Series.mask
s = df.groupby('cat').cumcount().mod(2)
df['tags'] = s.mask(df['cat'].eq('B'), ~s.astype(bool)).astype(int)
print(df)
cat tags
0 A 0
1 A 1
2 B 1
3 B 0
4 B 1
5 A 0
6 A 1
7 A 0
8 A 1

Answer 3

numpy place might help here :
#create temp column :
mapp={'A':0,'B':1}
res = (master.assign(temp=master.cat.map(mapp),
tag = master.cat
)
)
#locate point where B changes to A
split_point = res.loc[res.temp.diff().eq(-1)].index
split_point
Int64Index([5], dtype='int64')
#split into sections :
spl = np.split(res.cat,split_point)
def replace(entry):
np.place(entry.to_numpy(), entry=="A",[0,1])
np.place(entry.to_numpy(),entry=="B",[1,0])
return entry
res.tag = pd.concat(map(replace,spl))
res.drop('temp',axis=1)
cat tag
0 A 0
1 A 1
2 B 1
3 B 0
4 B 1
5 A 0
6 A 1
7 A 0
8 A 1
9 B 1
10 B 0
11 B 1
12 B 0
13 B 1