Mixing
Convert Columns to Rows every nth value in pandas and join with a comma
0
Here is what I have:
data = np.array([np.arange(1000)]*2).T
df = pd.DataFrame(data,columns=['a','b'])
a b
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
... ... ...
995 995 995
996 996 996
997 997 997
998 998 998
999 999 999
Here is what I am trying to do:
For column a, take the first 100th rows, append each row with a comma, add them as a row of their own. Take the next 100 rows. Continue till you exhaust the points in the column.
Repeat for column b
Overwriting the existing column, or creating a new column does not matter.
Here is what I am trying to make it look like:
c d
0,1,2,3,4,5...99 0,1,2,3,4,5...99
101, 102, 103,... 101, 102, 103,...
python pandas
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
add a comment |
0
Here is what I have:
data = np.array([np.arange(1000)]*2).T
df = pd.DataFrame(data,columns=['a','b'])
a b
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
... ... ...
995 995 995
996 996 996
997 997 997
998 998 998
999 999 999
Here is what I am trying to do:
For column a, take the first 100th rows, append each row with a comma, add them as a row of their own. Take the next 100 rows. Continue till you exhaust the points in the column.
Repeat for column b
Overwriting the existing column, or creating a new column does not matter.
Here is what I am trying to make it look like:
c d
0,1,2,3,4,5...99 0,1,2,3,4,5...99
101, 102, 103,... 101, 102, 103,...
python pandas
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
2
Doesdf.astype(str).groupby(df.index//100).agg(', '.join)do what you want?
– Jon Clements♦
Nov 21 '18 at 15:42
@JonClements yes this works to an extent, but breaks if the number of rows is 110 for example. Can it be robust against the number of rows? so it can work for 110, one row will have 100, the next will have 10?
– Matthew
Nov 21 '18 at 15:48
@JonClements sorry! Disregard my comment for a second !
– Matthew
Nov 21 '18 at 15:50
@JonClements yes this works perfectly ! Thank you. Please add it as an answer
– Matthew
Nov 21 '18 at 15:56
add a comment |
0
0
0
Here is what I have:
data = np.array([np.arange(1000)]*2).T
df = pd.DataFrame(data,columns=['a','b'])
a b
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
... ... ...
995 995 995
996 996 996
997 997 997
998 998 998
999 999 999
Here is what I am trying to do:
For column a, take the first 100th rows, append each row with a comma, add them as a row of their own. Take the next 100 rows. Continue till you exhaust the points in the column.
Repeat for column b
Overwriting the existing column, or creating a new column does not matter.
Here is what I am trying to make it look like:
c d
0,1,2,3,4,5...99 0,1,2,3,4,5...99
101, 102, 103,... 101, 102, 103,...
python pandas
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
Here is what I have:
data = np.array([np.arange(1000)]*2).T
df = pd.DataFrame(data,columns=['a','b'])
a b
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
... ... ...
995 995 995
996 996 996
997 997 997
998 998 998
999 999 999
Here is what I am trying to do:
For column a, take the first 100th rows, append each row with a comma, add them as a row of their own. Take the next 100 rows. Continue till you exhaust the points in the column.
Repeat for column b
Overwriting the existing column, or creating a new column does not matter.
Here is what I am trying to make it look like:
c d
0,1,2,3,4,5...99 0,1,2,3,4,5...99
101, 102, 103,... 101, 102, 103,...
python pandas
python pandas
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
asked Nov 21 '18 at 15:37
MatthewMatthew
89114
89114
2
Doesdf.astype(str).groupby(df.index//100).agg(', '.join)do what you want?
– Jon Clements♦
Nov 21 '18 at 15:42
@JonClements yes this works to an extent, but breaks if the number of rows is 110 for example. Can it be robust against the number of rows? so it can work for 110, one row will have 100, the next will have 10?
– Matthew
Nov 21 '18 at 15:48
@JonClements sorry! Disregard my comment for a second !
– Matthew
Nov 21 '18 at 15:50
@JonClements yes this works perfectly ! Thank you. Please add it as an answer
– Matthew
Nov 21 '18 at 15:56
add a comment |
2
Doesdf.astype(str).groupby(df.index//100).agg(', '.join)do what you want?
– Jon Clements♦
Nov 21 '18 at 15:42
@JonClements yes this works to an extent, but breaks if the number of rows is 110 for example. Can it be robust against the number of rows? so it can work for 110, one row will have 100, the next will have 10?
– Matthew
Nov 21 '18 at 15:48
@JonClements sorry! Disregard my comment for a second !
– Matthew
Nov 21 '18 at 15:50
@JonClements yes this works perfectly ! Thank you. Please add it as an answer
– Matthew
Nov 21 '18 at 15:56
2
2
Does
df.astype(str).groupby(df.index//100).agg(', '.join) do what you want?– Jon Clements♦
Nov 21 '18 at 15:42
Does
df.astype(str).groupby(df.index//100).agg(', '.join) do what you want?– Jon Clements♦
Nov 21 '18 at 15:42
@JonClements yes this works to an extent, but breaks if the number of rows is 110 for example. Can it be robust against the number of rows? so it can work for 110, one row will have 100, the next will have 10?
– Matthew
Nov 21 '18 at 15:48
@JonClements yes this works to an extent, but breaks if the number of rows is 110 for example. Can it be robust against the number of rows? so it can work for 110, one row will have 100, the next will have 10?
– Matthew
Nov 21 '18 at 15:48
@JonClements sorry! Disregard my comment for a second !
– Matthew
Nov 21 '18 at 15:50
@JonClements sorry! Disregard my comment for a second !
– Matthew
Nov 21 '18 at 15:50
@JonClements yes this works perfectly ! Thank you. Please add it as an answer
– Matthew
Nov 21 '18 at 15:56
@JonClements yes this works perfectly ! Thank you. Please add it as an answer
– Matthew
Nov 21 '18 at 15:56
add a comment |
1 Answer
1
active
oldest
votes
1
Here's something you could do:
df = df.assign(cut=pd.cut(x=df.a, bins = np.arange(0, len(data)+2, 100)))
And for a you could do:
df['c'] = df.groupby('cut')['a'].transform(lambda x: ','.join(map(str, x.values.tolist())))
df.drop('cut', axis = 1)
a b c
0 1 1 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
1 2 2 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
2 3 3 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
3 4 4 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
4 5 5 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Here's something you could do:
df = df.assign(cut=pd.cut(x=df.a, bins = np.arange(0, len(data)+2, 100)))
And for a you could do:
df['c'] = df.groupby('cut')['a'].transform(lambda x: ','.join(map(str, x.values.tolist())))
df.drop('cut', axis = 1)
a b c
0 1 1 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
1 2 2 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
2 3 3 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
3 4 4 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
4 5 5 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
add a comment |
1
Here's something you could do:
df = df.assign(cut=pd.cut(x=df.a, bins = np.arange(0, len(data)+2, 100)))
And for a you could do:
df['c'] = df.groupby('cut')['a'].transform(lambda x: ','.join(map(str, x.values.tolist())))
df.drop('cut', axis = 1)
a b c
0 1 1 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
1 2 2 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
2 3 3 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
3 4 4 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
4 5 5 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
add a comment |
1
1
1
Here's something you could do:
df = df.assign(cut=pd.cut(x=df.a, bins = np.arange(0, len(data)+2, 100)))
And for a you could do:
df['c'] = df.groupby('cut')['a'].transform(lambda x: ','.join(map(str, x.values.tolist())))
df.drop('cut', axis = 1)
a b c
0 1 1 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
1 2 2 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
2 3 3 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
3 4 4 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
4 5 5 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
Here's something you could do:
df = df.assign(cut=pd.cut(x=df.a, bins = np.arange(0, len(data)+2, 100)))
And for a you could do:
df['c'] = df.groupby('cut')['a'].transform(lambda x: ','.join(map(str, x.values.tolist())))
df.drop('cut', axis = 1)
a b c
0 1 1 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
1 2 2 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
2 3 3 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
3 4 4 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
4 5 5 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,1...
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
answered Nov 21 '18 at 16:10
yatuyatu
9,55611032
9,55611032
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
add a comment |
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
Albeit if @JonClements answer does work it is a more straight forward way of doing it
– yatu
Nov 21 '18 at 16:13
add a comment |
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2
Does
df.astype(str).groupby(df.index//100).agg(', '.join)do what you want?– Jon Clements♦
Nov 21 '18 at 15:42
@JonClements yes this works to an extent, but breaks if the number of rows is 110 for example. Can it be robust against the number of rows? so it can work for 110, one row will have 100, the next will have 10?
– Matthew
Nov 21 '18 at 15:48
@JonClements sorry! Disregard my comment for a second !
– Matthew
Nov 21 '18 at 15:50
@JonClements yes this works perfectly ! Thank you. Please add it as an answer
– Matthew
Nov 21 '18 at 15:56