How to group a dataframe and summarize over subgroups of consecutive numbers in Python?
I have a dataframe with a column containing ids and other column containing numbers:
df1 = {'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]}
You may note that each Id has their correponding series of consecutive numbers in the column "Number". For example:
Id 400 contains a series of length 4 {1, 2, 3, 4} and another of length 2 {8, 9}
I´d like to obtain for each Id, the average length of their corresponding series.
In this example:
df2 = {'ID':[400, 500], 'avg_length':[3, 2]}
Any ideas will be much appreciated!
python pandas pandas-groupby group-summaries
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
add a comment |
I have a dataframe with a column containing ids and other column containing numbers:
df1 = {'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]}
You may note that each Id has their correponding series of consecutive numbers in the column "Number". For example:
Id 400 contains a series of length 4 {1, 2, 3, 4} and another of length 2 {8, 9}
I´d like to obtain for each Id, the average length of their corresponding series.
In this example:
df2 = {'ID':[400, 500], 'avg_length':[3, 2]}
Any ideas will be much appreciated!
python pandas pandas-groupby group-summaries
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
add a comment |
I have a dataframe with a column containing ids and other column containing numbers:
df1 = {'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]}
You may note that each Id has their correponding series of consecutive numbers in the column "Number". For example:
Id 400 contains a series of length 4 {1, 2, 3, 4} and another of length 2 {8, 9}
I´d like to obtain for each Id, the average length of their corresponding series.
In this example:
df2 = {'ID':[400, 500], 'avg_length':[3, 2]}
Any ideas will be much appreciated!
python pandas pandas-groupby group-summaries
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
I have a dataframe with a column containing ids and other column containing numbers:
df1 = {'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]}
You may note that each Id has their correponding series of consecutive numbers in the column "Number". For example:
Id 400 contains a series of length 4 {1, 2, 3, 4} and another of length 2 {8, 9}
I´d like to obtain for each Id, the average length of their corresponding series.
In this example:
df2 = {'ID':[400, 500], 'avg_length':[3, 2]}
Any ideas will be much appreciated!
python pandas pandas-groupby group-summaries
python pandas pandas-groupby group-summaries
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
asked Nov 21 '18 at 16:29
Facundo IannelloFacundo Iannello
182
182
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Here is one way, uses groupby twice,
df1['tmp'] = (df1.Number - df1.Number.shift() > 1).cumsum()
df1.groupby(['ID', 'tmp']).Number.count().groupby(level = 0).mean().reset_index(name = 'avg_length')
2.29 ms ± 75.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
ID avg_length
0 400 3
1 500 2
Option 2: Without using apply twice, still uses tmp column created earlier
df1.groupby('ID').tmp.apply(lambda x: x.value_counts().mean()).reset_index(name = 'avg_length')
2.25 ms ± 99.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
add a comment |
groupby + cumsum + value_counts
You can use groupby with a custom function:
df = pd.DataFrame({'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]})
def mean_count(x):
return (x - x.shift()).ne(1).cumsum().value_counts().mean()
res = df.groupby('ID')['Number'].apply(mean_count).reset_index()
print(res)
ID Number
0 400 3.0
1 500 2.0
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
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
Here is one way, uses groupby twice,
df1['tmp'] = (df1.Number - df1.Number.shift() > 1).cumsum()
df1.groupby(['ID', 'tmp']).Number.count().groupby(level = 0).mean().reset_index(name = 'avg_length')
2.29 ms ± 75.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
ID avg_length
0 400 3
1 500 2
Option 2: Without using apply twice, still uses tmp column created earlier
df1.groupby('ID').tmp.apply(lambda x: x.value_counts().mean()).reset_index(name = 'avg_length')
2.25 ms ± 99.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
add a comment |
Here is one way, uses groupby twice,
df1['tmp'] = (df1.Number - df1.Number.shift() > 1).cumsum()
df1.groupby(['ID', 'tmp']).Number.count().groupby(level = 0).mean().reset_index(name = 'avg_length')
2.29 ms ± 75.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
ID avg_length
0 400 3
1 500 2
Option 2: Without using apply twice, still uses tmp column created earlier
df1.groupby('ID').tmp.apply(lambda x: x.value_counts().mean()).reset_index(name = 'avg_length')
2.25 ms ± 99.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
add a comment |
Here is one way, uses groupby twice,
df1['tmp'] = (df1.Number - df1.Number.shift() > 1).cumsum()
df1.groupby(['ID', 'tmp']).Number.count().groupby(level = 0).mean().reset_index(name = 'avg_length')
2.29 ms ± 75.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
ID avg_length
0 400 3
1 500 2
Option 2: Without using apply twice, still uses tmp column created earlier
df1.groupby('ID').tmp.apply(lambda x: x.value_counts().mean()).reset_index(name = 'avg_length')
2.25 ms ± 99.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
Here is one way, uses groupby twice,
df1['tmp'] = (df1.Number - df1.Number.shift() > 1).cumsum()
df1.groupby(['ID', 'tmp']).Number.count().groupby(level = 0).mean().reset_index(name = 'avg_length')
2.29 ms ± 75.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
ID avg_length
0 400 3
1 500 2
Option 2: Without using apply twice, still uses tmp column created earlier
df1.groupby('ID').tmp.apply(lambda x: x.value_counts().mean()).reset_index(name = 'avg_length')
2.25 ms ± 99.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
edited Nov 21 '18 at 17:00
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
answered Nov 21 '18 at 16:48
VaishaliVaishali
18.2k31029
18.2k31029
add a comment |
add a comment |
groupby + cumsum + value_counts
You can use groupby with a custom function:
df = pd.DataFrame({'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]})
def mean_count(x):
return (x - x.shift()).ne(1).cumsum().value_counts().mean()
res = df.groupby('ID')['Number'].apply(mean_count).reset_index()
print(res)
ID Number
0 400 3.0
1 500 2.0
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
add a comment |
groupby + cumsum + value_counts
You can use groupby with a custom function:
df = pd.DataFrame({'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]})
def mean_count(x):
return (x - x.shift()).ne(1).cumsum().value_counts().mean()
res = df.groupby('ID')['Number'].apply(mean_count).reset_index()
print(res)
ID Number
0 400 3.0
1 500 2.0
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
add a comment |
groupby + cumsum + value_counts
You can use groupby with a custom function:
df = pd.DataFrame({'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]})
def mean_count(x):
return (x - x.shift()).ne(1).cumsum().value_counts().mean()
res = df.groupby('ID')['Number'].apply(mean_count).reset_index()
print(res)
ID Number
0 400 3.0
1 500 2.0
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
groupby + cumsum + value_counts
You can use groupby with a custom function:
df = pd.DataFrame({'ID':[400, 400, 400, 400, 400, 400, 500, 500, 500, 500],
'Number':[1, 2, 3, 4, 8, 9, 22, 23, 26, 27]})
def mean_count(x):
return (x - x.shift()).ne(1).cumsum().value_counts().mean()
res = df.groupby('ID')['Number'].apply(mean_count).reset_index()
print(res)
ID Number
0 400 3.0
1 500 2.0
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
answered Nov 21 '18 at 16:56
jppjpp
93.8k2055106
93.8k2055106
add a comment |
add a comment |
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