python pandas consolidate rows with same values in a sequence and reorder (drop duplicates in a sequence)
up vote
1
down vote
favorite
Lets say i have following table:
ID FRUIT ORDER
01 apple 1
01 apple 2
01 peach 3
01 apple 4
02 melon 1
02 apple 2
02 apple 3
02 apple 4
Now i want to consolidate rows within same ID when the values are equal in a iterative manner (drop duplicates if they are in a sequence) and redefine the order number, e.g.
ID FRUIT ORDER
01 apple 1
01 peach 2
01 apple 3
02 melon 1
02 apple 2
EDIT: I forgot to reorder. Like above: the order should be re-arranged in an iterative manner
python pandas group-by
add a comment |
up vote
1
down vote
favorite
Lets say i have following table:
ID FRUIT ORDER
01 apple 1
01 apple 2
01 peach 3
01 apple 4
02 melon 1
02 apple 2
02 apple 3
02 apple 4
Now i want to consolidate rows within same ID when the values are equal in a iterative manner (drop duplicates if they are in a sequence) and redefine the order number, e.g.
ID FRUIT ORDER
01 apple 1
01 peach 2
01 apple 3
02 melon 1
02 apple 2
EDIT: I forgot to reorder. Like above: the order should be re-arranged in an iterative manner
python pandas group-by
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Lets say i have following table:
ID FRUIT ORDER
01 apple 1
01 apple 2
01 peach 3
01 apple 4
02 melon 1
02 apple 2
02 apple 3
02 apple 4
Now i want to consolidate rows within same ID when the values are equal in a iterative manner (drop duplicates if they are in a sequence) and redefine the order number, e.g.
ID FRUIT ORDER
01 apple 1
01 peach 2
01 apple 3
02 melon 1
02 apple 2
EDIT: I forgot to reorder. Like above: the order should be re-arranged in an iterative manner
python pandas group-by
Lets say i have following table:
ID FRUIT ORDER
01 apple 1
01 apple 2
01 peach 3
01 apple 4
02 melon 1
02 apple 2
02 apple 3
02 apple 4
Now i want to consolidate rows within same ID when the values are equal in a iterative manner (drop duplicates if they are in a sequence) and redefine the order number, e.g.
ID FRUIT ORDER
01 apple 1
01 peach 2
01 apple 3
02 melon 1
02 apple 2
EDIT: I forgot to reorder. Like above: the order should be re-arranged in an iterative manner
python pandas group-by
python pandas group-by
edited Nov 12 at 10:29
asked Nov 12 at 10:14
kxell2001
144
144
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Use boolean indexing
for filter only first consecutive values with cumcount
for new ordering:
a = df['ID'] + df['FRUIT']
#if necessary
#a = df['ID'].astype(str) + df['FRUIT']
df = df[a.ne(a.shift())]
df['ORDER'] = df.groupby('ID').cumcount().add(1)
print (df)
ID FRUIT ORDER
0 01 apple 1
2 01 peach 2
3 01 apple 3
4 02 melon 1
5 02 apple 2
add a comment |
up vote
0
down vote
I believe this will be easy one to go :
>>> df
ID FRUIT ORDER
0 01 apple 1
1 01 apple 2
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
6 02 apple 3
7 02 apple 4
>>> df[df['FRUIT'] != df['FRUIT'].shift(1)]
ID FRUIT ORDER
0 01 apple 1
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Use boolean indexing
for filter only first consecutive values with cumcount
for new ordering:
a = df['ID'] + df['FRUIT']
#if necessary
#a = df['ID'].astype(str) + df['FRUIT']
df = df[a.ne(a.shift())]
df['ORDER'] = df.groupby('ID').cumcount().add(1)
print (df)
ID FRUIT ORDER
0 01 apple 1
2 01 peach 2
3 01 apple 3
4 02 melon 1
5 02 apple 2
add a comment |
up vote
1
down vote
accepted
Use boolean indexing
for filter only first consecutive values with cumcount
for new ordering:
a = df['ID'] + df['FRUIT']
#if necessary
#a = df['ID'].astype(str) + df['FRUIT']
df = df[a.ne(a.shift())]
df['ORDER'] = df.groupby('ID').cumcount().add(1)
print (df)
ID FRUIT ORDER
0 01 apple 1
2 01 peach 2
3 01 apple 3
4 02 melon 1
5 02 apple 2
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Use boolean indexing
for filter only first consecutive values with cumcount
for new ordering:
a = df['ID'] + df['FRUIT']
#if necessary
#a = df['ID'].astype(str) + df['FRUIT']
df = df[a.ne(a.shift())]
df['ORDER'] = df.groupby('ID').cumcount().add(1)
print (df)
ID FRUIT ORDER
0 01 apple 1
2 01 peach 2
3 01 apple 3
4 02 melon 1
5 02 apple 2
Use boolean indexing
for filter only first consecutive values with cumcount
for new ordering:
a = df['ID'] + df['FRUIT']
#if necessary
#a = df['ID'].astype(str) + df['FRUIT']
df = df[a.ne(a.shift())]
df['ORDER'] = df.groupby('ID').cumcount().add(1)
print (df)
ID FRUIT ORDER
0 01 apple 1
2 01 peach 2
3 01 apple 3
4 02 melon 1
5 02 apple 2
edited Nov 12 at 15:38
answered Nov 12 at 10:26
jezrael
316k22256333
316k22256333
add a comment |
add a comment |
up vote
0
down vote
I believe this will be easy one to go :
>>> df
ID FRUIT ORDER
0 01 apple 1
1 01 apple 2
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
6 02 apple 3
7 02 apple 4
>>> df[df['FRUIT'] != df['FRUIT'].shift(1)]
ID FRUIT ORDER
0 01 apple 1
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
add a comment |
up vote
0
down vote
I believe this will be easy one to go :
>>> df
ID FRUIT ORDER
0 01 apple 1
1 01 apple 2
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
6 02 apple 3
7 02 apple 4
>>> df[df['FRUIT'] != df['FRUIT'].shift(1)]
ID FRUIT ORDER
0 01 apple 1
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
add a comment |
up vote
0
down vote
up vote
0
down vote
I believe this will be easy one to go :
>>> df
ID FRUIT ORDER
0 01 apple 1
1 01 apple 2
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
6 02 apple 3
7 02 apple 4
>>> df[df['FRUIT'] != df['FRUIT'].shift(1)]
ID FRUIT ORDER
0 01 apple 1
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
I believe this will be easy one to go :
>>> df
ID FRUIT ORDER
0 01 apple 1
1 01 apple 2
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
6 02 apple 3
7 02 apple 4
>>> df[df['FRUIT'] != df['FRUIT'].shift(1)]
ID FRUIT ORDER
0 01 apple 1
2 01 peach 3
3 01 apple 4
4 02 melon 1
5 02 apple 2
edited Nov 12 at 10:55
answered Nov 12 at 10:45
pygo
1,7361416
1,7361416
add a comment |
add a comment |
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