Pandas: How to obtain top 2, middle 2 and bottom 2 rows in a each group
Let's say I have a dataframe df as below. To obtain 1st 2 and last 2 in each group I have used groupby.nth
df = pd.DataFrame({'A': ['a','a','a','a','a','a','a','a','b','b','b','b','b','b','b'],
'B': [1, 2, 3, 4, 5,6,7,8,1, 2, 3, 4, 5,6,7]}, columns=['A', 'B'])
df.groupby('A').nth([0,1,-2,-1])
Result:
B
A
a 1
a 2
a 7
a 8
b 1
b 2
b 6
b 7
I'm not sure how to obtain the middle 2 rows. For example, in group 'A' there are 8 instances so my middle would be 4, 5 (n/2, n/2+1) and group 'B' my middle rows would be 3, 4 (n/2-0.5, n/2+0.5). Any guidance is appreciated.
pandas pandas-groupby
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
asked Nov 15 '18 at 3:33
psangampsangam
444
add a comment |
Let's say I have a dataframe df as below. To obtain 1st 2 and last 2 in each group I have used groupby.nth
df = pd.DataFrame({'A': ['a','a','a','a','a','a','a','a','b','b','b','b','b','b','b'],
'B': [1, 2, 3, 4, 5,6,7,8,1, 2, 3, 4, 5,6,7]}, columns=['A', 'B'])
df.groupby('A').nth([0,1,-2,-1])
Result:
B
A
a 1
a 2
a 7
a 8
b 1
b 2
b 6
b 7
I'm not sure how to obtain the middle 2 rows. For example, in group 'A' there are 8 instances so my middle would be 4, 5 (n/2, n/2+1) and group 'B' my middle rows would be 3, 4 (n/2-0.5, n/2+0.5). Any guidance is appreciated.
pandas pandas-groupby
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
asked Nov 15 '18 at 3:33
psangampsangam
444
add a comment |
Let's say I have a dataframe df as below. To obtain 1st 2 and last 2 in each group I have used groupby.nth
df = pd.DataFrame({'A': ['a','a','a','a','a','a','a','a','b','b','b','b','b','b','b'],
'B': [1, 2, 3, 4, 5,6,7,8,1, 2, 3, 4, 5,6,7]}, columns=['A', 'B'])
df.groupby('A').nth([0,1,-2,-1])
Result:
B
A
a 1
a 2
a 7
a 8
b 1
b 2
b 6
b 7
I'm not sure how to obtain the middle 2 rows. For example, in group 'A' there are 8 instances so my middle would be 4, 5 (n/2, n/2+1) and group 'B' my middle rows would be 3, 4 (n/2-0.5, n/2+0.5). Any guidance is appreciated.
pandas pandas-groupby
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
asked Nov 15 '18 at 3:33
psangampsangam
444
Let's say I have a dataframe df as below. To obtain 1st 2 and last 2 in each group I have used groupby.nth
df = pd.DataFrame({'A': ['a','a','a','a','a','a','a','a','b','b','b','b','b','b','b'],
'B': [1, 2, 3, 4, 5,6,7,8,1, 2, 3, 4, 5,6,7]}, columns=['A', 'B'])
df.groupby('A').nth([0,1,-2,-1])
Result:
B
A
a 1
a 2
a 7
a 8
b 1
b 2
b 6
b 7
I'm not sure how to obtain the middle 2 rows. For example, in group 'A' there are 8 instances so my middle would be 4, 5 (n/2, n/2+1) and group 'B' my middle rows would be 3, 4 (n/2-0.5, n/2+0.5). Any guidance is appreciated.
pandas pandas-groupby
pandas pandas-groupby
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
asked Nov 15 '18 at 3:33
psangampsangam
444
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
asked Nov 15 '18 at 3:33
psangampsangam
444
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
edited Nov 15 '18 at 6:23
Aqueous Carlos
374314
374314
asked Nov 15 '18 at 3:33
psangampsangam
444
asked Nov 15 '18 at 3:33
psangampsangam
444
asked Nov 15 '18 at 3:33
psangampsangam
444
444
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
sacul's answer is nice , Here I just follow your own idea def a customize function
def middle(x):
if len(x) % 2 == 0:
return x.iloc[int(len(x) / 2) - 1:int(len(x) / 2) + 1]
else:
return x.iloc[int((len(x) / 2 - 0.5)) - 1:int(len(x) / 2 + 0.5)]
pd.concat([middle(y) for _ , y in df.groupby('A')])
Out[25]:
A B
3 a 4
4 a 5
10 b 3
11 b 4
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
add a comment |
You can use iloc to find the n//2 -1 and n//2 indices for each group (// is floor division):
g = df.groupby('A')
g.apply(lambda x: x['B'].iloc[[len(x)//2-1, len(x)//2]])
A
a 3 4
4 5
b 10 3
11 4
Name: B, dtype: int64
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
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
sacul's answer is nice , Here I just follow your own idea def a customize function
def middle(x):
if len(x) % 2 == 0:
return x.iloc[int(len(x) / 2) - 1:int(len(x) / 2) + 1]
else:
return x.iloc[int((len(x) / 2 - 0.5)) - 1:int(len(x) / 2 + 0.5)]
pd.concat([middle(y) for _ , y in df.groupby('A')])
Out[25]:
A B
3 a 4
4 a 5
10 b 3
11 b 4
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
add a comment |
sacul's answer is nice , Here I just follow your own idea def a customize function
def middle(x):
if len(x) % 2 == 0:
return x.iloc[int(len(x) / 2) - 1:int(len(x) / 2) + 1]
else:
return x.iloc[int((len(x) / 2 - 0.5)) - 1:int(len(x) / 2 + 0.5)]
pd.concat([middle(y) for _ , y in df.groupby('A')])
Out[25]:
A B
3 a 4
4 a 5
10 b 3
11 b 4
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
add a comment |
sacul's answer is nice , Here I just follow your own idea def a customize function
def middle(x):
if len(x) % 2 == 0:
return x.iloc[int(len(x) / 2) - 1:int(len(x) / 2) + 1]
else:
return x.iloc[int((len(x) / 2 - 0.5)) - 1:int(len(x) / 2 + 0.5)]
pd.concat([middle(y) for _ , y in df.groupby('A')])
Out[25]:
A B
3 a 4
4 a 5
10 b 3
11 b 4
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
sacul's answer is nice , Here I just follow your own idea def a customize function
def middle(x):
if len(x) % 2 == 0:
return x.iloc[int(len(x) / 2) - 1:int(len(x) / 2) + 1]
else:
return x.iloc[int((len(x) / 2 - 0.5)) - 1:int(len(x) / 2 + 0.5)]
pd.concat([middle(y) for _ , y in df.groupby('A')])
Out[25]:
A B
3 a 4
4 a 5
10 b 3
11 b 4
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
answered Nov 15 '18 at 3:51
Wen-BenWen-Ben
114k83368
114k83368
add a comment |
add a comment |
You can use iloc to find the n//2 -1 and n//2 indices for each group (// is floor division):
g = df.groupby('A')
g.apply(lambda x: x['B'].iloc[[len(x)//2-1, len(x)//2]])
A
a 3 4
4 5
b 10 3
11 4
Name: B, dtype: int64
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
add a comment |
You can use iloc to find the n//2 -1 and n//2 indices for each group (// is floor division):
g = df.groupby('A')
g.apply(lambda x: x['B'].iloc[[len(x)//2-1, len(x)//2]])
A
a 3 4
4 5
b 10 3
11 4
Name: B, dtype: int64
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
add a comment |
You can use iloc to find the n//2 -1 and n//2 indices for each group (// is floor division):
g = df.groupby('A')
g.apply(lambda x: x['B'].iloc[[len(x)//2-1, len(x)//2]])
A
a 3 4
4 5
b 10 3
11 4
Name: B, dtype: int64
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
You can use iloc to find the n//2 -1 and n//2 indices for each group (// is floor division):
g = df.groupby('A')
g.apply(lambda x: x['B'].iloc[[len(x)//2-1, len(x)//2]])
A
a 3 4
4 5
b 10 3
11 4
Name: B, dtype: int64
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
edited Nov 15 '18 at 3:55
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
answered Nov 15 '18 at 3:48
sacuLsacuL
30.5k41942
30.5k41942
add a comment |
add a comment |
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