Pandas Series value_counts working differently for different counts
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For example:
df1 = pd.DataFrame(np.repeat(np.arange(1,7),3), columns=['A'])
df1.A.value_counts(sort=False)
1 3
2 3
3 3
4 3
5 3
6 3
Name: A, dtype: int64
df2 = pd.DataFrame(np.repeat(np.arange(1,7),100), columns=['A'])
df2.A.value_counts(sort=False)
1 100
2 100
3 100
4 100
5 100
6 100
Name: A, dtype: int64
In the above examples the value_counts
works perfectly and give the required result. whereas when coming to larger dataframes it is giving a different output. Here the A
values are already sorted and counts are also same, but the order of index that is A
changed after value_counts
. Why is it doing correctly for small counts but not for large counts:
df3 = pd.DataFrame(np.repeat(np.arange(1,7),1000), columns=['A'])
df3.A.value_counts(sort=False)
4 1000
1 1000
5 1000
2 1000
6 1000
3 1000
Name: A, dtype: int64
Here I can do df3.A.value_counts(sort=False).sort_index()
or df3.A.value_counts(sort=False).reindex(df.A.unique())
. I want to know the reason why it is behaving differently for different counts?
Using:
Numpy version :1.15.2
Pandas version :0.23.4
python pandas numpy
add a comment |
For example:
df1 = pd.DataFrame(np.repeat(np.arange(1,7),3), columns=['A'])
df1.A.value_counts(sort=False)
1 3
2 3
3 3
4 3
5 3
6 3
Name: A, dtype: int64
df2 = pd.DataFrame(np.repeat(np.arange(1,7),100), columns=['A'])
df2.A.value_counts(sort=False)
1 100
2 100
3 100
4 100
5 100
6 100
Name: A, dtype: int64
In the above examples the value_counts
works perfectly and give the required result. whereas when coming to larger dataframes it is giving a different output. Here the A
values are already sorted and counts are also same, but the order of index that is A
changed after value_counts
. Why is it doing correctly for small counts but not for large counts:
df3 = pd.DataFrame(np.repeat(np.arange(1,7),1000), columns=['A'])
df3.A.value_counts(sort=False)
4 1000
1 1000
5 1000
2 1000
6 1000
3 1000
Name: A, dtype: int64
Here I can do df3.A.value_counts(sort=False).sort_index()
or df3.A.value_counts(sort=False).reindex(df.A.unique())
. I want to know the reason why it is behaving differently for different counts?
Using:
Numpy version :1.15.2
Pandas version :0.23.4
python pandas numpy
add a comment |
For example:
df1 = pd.DataFrame(np.repeat(np.arange(1,7),3), columns=['A'])
df1.A.value_counts(sort=False)
1 3
2 3
3 3
4 3
5 3
6 3
Name: A, dtype: int64
df2 = pd.DataFrame(np.repeat(np.arange(1,7),100), columns=['A'])
df2.A.value_counts(sort=False)
1 100
2 100
3 100
4 100
5 100
6 100
Name: A, dtype: int64
In the above examples the value_counts
works perfectly and give the required result. whereas when coming to larger dataframes it is giving a different output. Here the A
values are already sorted and counts are also same, but the order of index that is A
changed after value_counts
. Why is it doing correctly for small counts but not for large counts:
df3 = pd.DataFrame(np.repeat(np.arange(1,7),1000), columns=['A'])
df3.A.value_counts(sort=False)
4 1000
1 1000
5 1000
2 1000
6 1000
3 1000
Name: A, dtype: int64
Here I can do df3.A.value_counts(sort=False).sort_index()
or df3.A.value_counts(sort=False).reindex(df.A.unique())
. I want to know the reason why it is behaving differently for different counts?
Using:
Numpy version :1.15.2
Pandas version :0.23.4
python pandas numpy
For example:
df1 = pd.DataFrame(np.repeat(np.arange(1,7),3), columns=['A'])
df1.A.value_counts(sort=False)
1 3
2 3
3 3
4 3
5 3
6 3
Name: A, dtype: int64
df2 = pd.DataFrame(np.repeat(np.arange(1,7),100), columns=['A'])
df2.A.value_counts(sort=False)
1 100
2 100
3 100
4 100
5 100
6 100
Name: A, dtype: int64
In the above examples the value_counts
works perfectly and give the required result. whereas when coming to larger dataframes it is giving a different output. Here the A
values are already sorted and counts are also same, but the order of index that is A
changed after value_counts
. Why is it doing correctly for small counts but not for large counts:
df3 = pd.DataFrame(np.repeat(np.arange(1,7),1000), columns=['A'])
df3.A.value_counts(sort=False)
4 1000
1 1000
5 1000
2 1000
6 1000
3 1000
Name: A, dtype: int64
Here I can do df3.A.value_counts(sort=False).sort_index()
or df3.A.value_counts(sort=False).reindex(df.A.unique())
. I want to know the reason why it is behaving differently for different counts?
Using:
Numpy version :1.15.2
Pandas version :0.23.4
python pandas numpy
python pandas numpy
edited Nov 14 '18 at 6:29
Sandeep Kadapa
asked Nov 14 '18 at 6:22
Sandeep KadapaSandeep Kadapa
7,423831
7,423831
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This is actually a known problem.
If you browse through the source code -
C:ProgramDataAnaconda3Libsite-packagespandascorealgorithims.py
line581
is the original implementation- It calls
_value_counts_arraylike
forint64
values whenbins=None
- This function makes a call -
keys, counts = htable.value_count_int64(values, dropna)
If you then look at the htable
implementation you will conclude that the keys are in an arbitrary order, subject to how the hashtable
works.
Its not a guarantee of ANY kind of ordering. Typically this routine sorts by biggest values, and that is almost always what you want.
I guess they can change this to have sort=False
mean original ordering. I don't know if this would actually break anything (and done internally this isn't very costly as the uniques are already known).
The order is changed from pandas/hashtable.pyx.build_count_table_object()
. Resizing of the pymap
moves the entries by hashing values.
Here is the full discussion
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
1
It looks to be some update for orignal order retaining github page look to be using same update asreindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering
– pygo
Nov 14 '18 at 8:16
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
This is actually a known problem.
If you browse through the source code -
C:ProgramDataAnaconda3Libsite-packagespandascorealgorithims.py
line581
is the original implementation- It calls
_value_counts_arraylike
forint64
values whenbins=None
- This function makes a call -
keys, counts = htable.value_count_int64(values, dropna)
If you then look at the htable
implementation you will conclude that the keys are in an arbitrary order, subject to how the hashtable
works.
Its not a guarantee of ANY kind of ordering. Typically this routine sorts by biggest values, and that is almost always what you want.
I guess they can change this to have sort=False
mean original ordering. I don't know if this would actually break anything (and done internally this isn't very costly as the uniques are already known).
The order is changed from pandas/hashtable.pyx.build_count_table_object()
. Resizing of the pymap
moves the entries by hashing values.
Here is the full discussion
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
1
It looks to be some update for orignal order retaining github page look to be using same update asreindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering
– pygo
Nov 14 '18 at 8:16
add a comment |
This is actually a known problem.
If you browse through the source code -
C:ProgramDataAnaconda3Libsite-packagespandascorealgorithims.py
line581
is the original implementation- It calls
_value_counts_arraylike
forint64
values whenbins=None
- This function makes a call -
keys, counts = htable.value_count_int64(values, dropna)
If you then look at the htable
implementation you will conclude that the keys are in an arbitrary order, subject to how the hashtable
works.
Its not a guarantee of ANY kind of ordering. Typically this routine sorts by biggest values, and that is almost always what you want.
I guess they can change this to have sort=False
mean original ordering. I don't know if this would actually break anything (and done internally this isn't very costly as the uniques are already known).
The order is changed from pandas/hashtable.pyx.build_count_table_object()
. Resizing of the pymap
moves the entries by hashing values.
Here is the full discussion
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
1
It looks to be some update for orignal order retaining github page look to be using same update asreindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering
– pygo
Nov 14 '18 at 8:16
add a comment |
This is actually a known problem.
If you browse through the source code -
C:ProgramDataAnaconda3Libsite-packagespandascorealgorithims.py
line581
is the original implementation- It calls
_value_counts_arraylike
forint64
values whenbins=None
- This function makes a call -
keys, counts = htable.value_count_int64(values, dropna)
If you then look at the htable
implementation you will conclude that the keys are in an arbitrary order, subject to how the hashtable
works.
Its not a guarantee of ANY kind of ordering. Typically this routine sorts by biggest values, and that is almost always what you want.
I guess they can change this to have sort=False
mean original ordering. I don't know if this would actually break anything (and done internally this isn't very costly as the uniques are already known).
The order is changed from pandas/hashtable.pyx.build_count_table_object()
. Resizing of the pymap
moves the entries by hashing values.
Here is the full discussion
This is actually a known problem.
If you browse through the source code -
C:ProgramDataAnaconda3Libsite-packagespandascorealgorithims.py
line581
is the original implementation- It calls
_value_counts_arraylike
forint64
values whenbins=None
- This function makes a call -
keys, counts = htable.value_count_int64(values, dropna)
If you then look at the htable
implementation you will conclude that the keys are in an arbitrary order, subject to how the hashtable
works.
Its not a guarantee of ANY kind of ordering. Typically this routine sorts by biggest values, and that is almost always what you want.
I guess they can change this to have sort=False
mean original ordering. I don't know if this would actually break anything (and done internally this isn't very costly as the uniques are already known).
The order is changed from pandas/hashtable.pyx.build_count_table_object()
. Resizing of the pymap
moves the entries by hashing values.
Here is the full discussion
answered Nov 14 '18 at 6:44
Vivek KalyanaranganVivek Kalyanarangan
5,1141830
5,1141830
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
1
It looks to be some update for orignal order retaining github page look to be using same update asreindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering
– pygo
Nov 14 '18 at 8:16
add a comment |
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
1
It looks to be some update for orignal order retaining github page look to be using same update asreindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering
– pygo
Nov 14 '18 at 8:16
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
When keys are in an arbitrary order, it must not guarantee for any number of counts right? but if the counts are small it is preserving the order and if the counts are high it isn't. Also, it maintains order till 341 counts but fails after it.
– Sandeep Kadapa
Nov 14 '18 at 7:05
1
1
It looks to be some update for orignal order retaining github page look to be using same update as
reindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering– pygo
Nov 14 '18 at 8:16
It looks to be some update for orignal order retaining github page look to be using same update as
reindex(unique(values))
:-) that's only way, So one have to reindex to preseve the orignal ordering– pygo
Nov 14 '18 at 8:16
add a comment |
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