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Group by consecutive index numbers
Finding the index of an item given a list containing it in PythonAccessing the index in 'for' loops?How do I get the number of elements in a list?Using group by on multiple columnsSelect first row in each GROUP BY group?Group by in LINQUnicodeEncodeError: 'ascii' codec can't encode character u'xa0' in position 20: ordinal not in range(128)Adding new column to existing DataFrame in Python pandas“Large data” work flows using pandasApply multiple functions to multiple groupby columns
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
I was wondering if there is a way to groupby consecutive index numbers and move the groups in different columns. Here is an example of the DataFrame I'm using:
0
0 19218.965703
1 19247.621650
2 19232.651322
9 19279.216956
10 19330.087371
11 19304.316973
And my idea is to gruoup by sequential index numbers and get something like this:
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Ive been trying to split my data by blocks of 3 and then groupby but I was looking more about something that can be used to group and rearrange sequential index numbers.
Thank you!
python pandas numpy group-by
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
add a comment |
I was wondering if there is a way to groupby consecutive index numbers and move the groups in different columns. Here is an example of the DataFrame I'm using:
0
0 19218.965703
1 19247.621650
2 19232.651322
9 19279.216956
10 19330.087371
11 19304.316973
And my idea is to gruoup by sequential index numbers and get something like this:
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Ive been trying to split my data by blocks of 3 and then groupby but I was looking more about something that can be used to group and rearrange sequential index numbers.
Thank you!
python pandas numpy group-by
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
add a comment |
I was wondering if there is a way to groupby consecutive index numbers and move the groups in different columns. Here is an example of the DataFrame I'm using:
0
0 19218.965703
1 19247.621650
2 19232.651322
9 19279.216956
10 19330.087371
11 19304.316973
And my idea is to gruoup by sequential index numbers and get something like this:
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Ive been trying to split my data by blocks of 3 and then groupby but I was looking more about something that can be used to group and rearrange sequential index numbers.
Thank you!
python pandas numpy group-by
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
I was wondering if there is a way to groupby consecutive index numbers and move the groups in different columns. Here is an example of the DataFrame I'm using:
0
0 19218.965703
1 19247.621650
2 19232.651322
9 19279.216956
10 19330.087371
11 19304.316973
And my idea is to gruoup by sequential index numbers and get something like this:
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Ive been trying to split my data by blocks of 3 and then groupby but I was looking more about something that can be used to group and rearrange sequential index numbers.
Thank you!
python pandas numpy group-by
python pandas numpy group-by
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
asked 9 hours ago
GiuseppeGiuseppe
1016 bronze badges
1016 bronze badges
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Here is one way:
from more_itertools import consecutive_groups
final=pd.concat([df.loc[i].reset_index(drop=True)
for i in consecutive_groups(df.index)],axis=1)
final.columns=range(len(final.columns))
print(final)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
1
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
add a comment |
This is a groupby + pivot_table
m = df.index.to_series().diff().ne(1).cumsum()
(df.assign(key=df.groupby(m).cumcount())
.pivot_table(index='key', columns=m, values=0))
1 2
key
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
add a comment |
Create a new pandas.Series with a new pandas.MultiIndex
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
pd.Series(df['0'].to_numpy(), [b, a]).unstack()
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Similar but with more Numpy
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
c = np.empty((b.max() + 1, a.max() + 1), float)
c.fill(np.nan)
c[b, a] = np.ravel(df)
pd.DataFrame(c)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
add a comment |
One way from pandas groupby
s=df.index.to_series().diff().ne(1).cumsum()
pd.concat({x: y.reset_index(drop=True) for x, y in df['0'].groupby(s)}, axis=1)
Out[786]:
1 2
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is one way:
from more_itertools import consecutive_groups
final=pd.concat([df.loc[i].reset_index(drop=True)
for i in consecutive_groups(df.index)],axis=1)
final.columns=range(len(final.columns))
print(final)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
1
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
add a comment |
Here is one way:
from more_itertools import consecutive_groups
final=pd.concat([df.loc[i].reset_index(drop=True)
for i in consecutive_groups(df.index)],axis=1)
final.columns=range(len(final.columns))
print(final)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
1
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
add a comment |
Here is one way:
from more_itertools import consecutive_groups
final=pd.concat([df.loc[i].reset_index(drop=True)
for i in consecutive_groups(df.index)],axis=1)
final.columns=range(len(final.columns))
print(final)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
Here is one way:
from more_itertools import consecutive_groups
final=pd.concat([df.loc[i].reset_index(drop=True)
for i in consecutive_groups(df.index)],axis=1)
final.columns=range(len(final.columns))
print(final)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
answered 9 hours ago
anky_91anky_91
23.8k5 gold badges12 silver badges29 bronze badges
23.8k5 gold badges12 silver badges29 bronze badges
1
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
add a comment |
1
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
1
1
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
I like the more_itertools solution! Thank you. With 3 answers you guys covered all the possible and elegant solutions!!
– Giuseppe
9 hours ago
add a comment |
This is a groupby + pivot_table
m = df.index.to_series().diff().ne(1).cumsum()
(df.assign(key=df.groupby(m).cumcount())
.pivot_table(index='key', columns=m, values=0))
1 2
key
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
add a comment |
This is a groupby + pivot_table
m = df.index.to_series().diff().ne(1).cumsum()
(df.assign(key=df.groupby(m).cumcount())
.pivot_table(index='key', columns=m, values=0))
1 2
key
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
add a comment |
This is a groupby + pivot_table
m = df.index.to_series().diff().ne(1).cumsum()
(df.assign(key=df.groupby(m).cumcount())
.pivot_table(index='key', columns=m, values=0))
1 2
key
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
This is a groupby + pivot_table
m = df.index.to_series().diff().ne(1).cumsum()
(df.assign(key=df.groupby(m).cumcount())
.pivot_table(index='key', columns=m, values=0))
1 2
key
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
answered 9 hours ago
user3483203user3483203
38.1k8 gold badges32 silver badges62 bronze badges
38.1k8 gold badges32 silver badges62 bronze badges
add a comment |
add a comment |
Create a new pandas.Series with a new pandas.MultiIndex
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
pd.Series(df['0'].to_numpy(), [b, a]).unstack()
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Similar but with more Numpy
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
c = np.empty((b.max() + 1, a.max() + 1), float)
c.fill(np.nan)
c[b, a] = np.ravel(df)
pd.DataFrame(c)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
add a comment |
Create a new pandas.Series with a new pandas.MultiIndex
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
pd.Series(df['0'].to_numpy(), [b, a]).unstack()
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Similar but with more Numpy
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
c = np.empty((b.max() + 1, a.max() + 1), float)
c.fill(np.nan)
c[b, a] = np.ravel(df)
pd.DataFrame(c)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
add a comment |
Create a new pandas.Series with a new pandas.MultiIndex
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
pd.Series(df['0'].to_numpy(), [b, a]).unstack()
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Similar but with more Numpy
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
c = np.empty((b.max() + 1, a.max() + 1), float)
c.fill(np.nan)
c[b, a] = np.ravel(df)
pd.DataFrame(c)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
Create a new pandas.Series with a new pandas.MultiIndex
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
pd.Series(df['0'].to_numpy(), [b, a]).unstack()
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
Similar but with more Numpy
a = pd.factorize(df.index - np.arange(len(df)))[0]
b = df.groupby(a).cumcount()
c = np.empty((b.max() + 1, a.max() + 1), float)
c.fill(np.nan)
c[b, a] = np.ravel(df)
pd.DataFrame(c)
0 1
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
edited 9 hours ago
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
answered 9 hours ago
piRSquaredpiRSquared
177k26 gold badges195 silver badges352 bronze badges
177k26 gold badges195 silver badges352 bronze badges
add a comment |
add a comment |
One way from pandas groupby
s=df.index.to_series().diff().ne(1).cumsum()
pd.concat({x: y.reset_index(drop=True) for x, y in df['0'].groupby(s)}, axis=1)
Out[786]:
1 2
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
add a comment |
One way from pandas groupby
s=df.index.to_series().diff().ne(1).cumsum()
pd.concat({x: y.reset_index(drop=True) for x, y in df['0'].groupby(s)}, axis=1)
Out[786]:
1 2
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
add a comment |
One way from pandas groupby
s=df.index.to_series().diff().ne(1).cumsum()
pd.concat({x: y.reset_index(drop=True) for x, y in df['0'].groupby(s)}, axis=1)
Out[786]:
1 2
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
One way from pandas groupby
s=df.index.to_series().diff().ne(1).cumsum()
pd.concat({x: y.reset_index(drop=True) for x, y in df['0'].groupby(s)}, axis=1)
Out[786]:
1 2
0 19218.965703 19279.216956
1 19247.621650 19330.087371
2 19232.651322 19304.316973
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
answered 9 hours ago
WeNYoBenWeNYoBen
156k8 gold badges54 silver badges84 bronze badges
156k8 gold badges54 silver badges84 bronze badges
add a comment |
add a comment |
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