Get overall smallest elements' distribution in dataframe with sorted columns more efficiently
I have a dataframe with sorted columns, something like this:
df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2)) for q in ['blue', 'green', 'red'])
blue green red
0 -2.15 -0.76 -2.62
1 -0.88 -0.62 -1.65
2 -0.77 -0.55 -1.51
3 -0.73 -0.17 -1.14
4 -0.06 -0.16 -0.75
5 -0.03 0.05 -0.08
6 0.06 0.38 0.37
7 0.41 0.76 1.04
8 0.56 0.89 1.16
9 0.97 2.94 1.79
What I want to know is how many of the n smallest elements in the whole frame are in each column. This is the only thing I came up with:
is_small = df.isin(np.partition(df.values.flatten(), n)[:n])
with n=10 it looks like this:
blue green red
0 True True True
1 True False True
2 True False True
3 True False True
4 False False True
5 False False False
6 False False False
7 False False False
8 False False False
9 False False False
Then by applying np.sum I get the number corresponding to each column.
I'm dissatisfied with this solution because it in no way utilizes the sortedness of the original data. All the data gets partitioned and all the data is then checked for whether it's in the partition. It seems wasteful, and I can't seem to figure out a better way.
python python-3.x pandas numpy
add a comment |
I have a dataframe with sorted columns, something like this:
df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2)) for q in ['blue', 'green', 'red'])
blue green red
0 -2.15 -0.76 -2.62
1 -0.88 -0.62 -1.65
2 -0.77 -0.55 -1.51
3 -0.73 -0.17 -1.14
4 -0.06 -0.16 -0.75
5 -0.03 0.05 -0.08
6 0.06 0.38 0.37
7 0.41 0.76 1.04
8 0.56 0.89 1.16
9 0.97 2.94 1.79
What I want to know is how many of the n smallest elements in the whole frame are in each column. This is the only thing I came up with:
is_small = df.isin(np.partition(df.values.flatten(), n)[:n])
with n=10 it looks like this:
blue green red
0 True True True
1 True False True
2 True False True
3 True False True
4 False False True
5 False False False
6 False False False
7 False False False
8 False False False
9 False False False
Then by applying np.sum I get the number corresponding to each column.
I'm dissatisfied with this solution because it in no way utilizes the sortedness of the original data. All the data gets partitioned and all the data is then checked for whether it's in the partition. It seems wasteful, and I can't seem to figure out a better way.
python python-3.x pandas numpy
add a comment |
I have a dataframe with sorted columns, something like this:
df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2)) for q in ['blue', 'green', 'red'])
blue green red
0 -2.15 -0.76 -2.62
1 -0.88 -0.62 -1.65
2 -0.77 -0.55 -1.51
3 -0.73 -0.17 -1.14
4 -0.06 -0.16 -0.75
5 -0.03 0.05 -0.08
6 0.06 0.38 0.37
7 0.41 0.76 1.04
8 0.56 0.89 1.16
9 0.97 2.94 1.79
What I want to know is how many of the n smallest elements in the whole frame are in each column. This is the only thing I came up with:
is_small = df.isin(np.partition(df.values.flatten(), n)[:n])
with n=10 it looks like this:
blue green red
0 True True True
1 True False True
2 True False True
3 True False True
4 False False True
5 False False False
6 False False False
7 False False False
8 False False False
9 False False False
Then by applying np.sum I get the number corresponding to each column.
I'm dissatisfied with this solution because it in no way utilizes the sortedness of the original data. All the data gets partitioned and all the data is then checked for whether it's in the partition. It seems wasteful, and I can't seem to figure out a better way.
python python-3.x pandas numpy
I have a dataframe with sorted columns, something like this:
df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2)) for q in ['blue', 'green', 'red'])
blue green red
0 -2.15 -0.76 -2.62
1 -0.88 -0.62 -1.65
2 -0.77 -0.55 -1.51
3 -0.73 -0.17 -1.14
4 -0.06 -0.16 -0.75
5 -0.03 0.05 -0.08
6 0.06 0.38 0.37
7 0.41 0.76 1.04
8 0.56 0.89 1.16
9 0.97 2.94 1.79
What I want to know is how many of the n smallest elements in the whole frame are in each column. This is the only thing I came up with:
is_small = df.isin(np.partition(df.values.flatten(), n)[:n])
with n=10 it looks like this:
blue green red
0 True True True
1 True False True
2 True False True
3 True False True
4 False False True
5 False False False
6 False False False
7 False False False
8 False False False
9 False False False
Then by applying np.sum I get the number corresponding to each column.
I'm dissatisfied with this solution because it in no way utilizes the sortedness of the original data. All the data gets partitioned and all the data is then checked for whether it's in the partition. It seems wasteful, and I can't seem to figure out a better way.
python python-3.x pandas numpy
python python-3.x pandas numpy
asked Nov 15 '18 at 19:17
MegaBluejayMegaBluejay
36618
36618
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Think you can compare the largest of n-smallest values against the partitioned one and then use idxmin
to leverage the sorted nature -
# Find largest of n smallest numbers
N = (np.partition(df.values.flatten(), n)[:n]).max()
out = (df<=N).idxmin(axis=0)
Sample run -
In [152]: np.random.seed(0)
In [153]: df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2))
for q in ['blue', 'green', 'red'])
In [154]: df
Out[154]:
blue green red
0 -0.98 -0.85 -2.55
1 -0.15 -0.21 -1.45
2 -0.10 0.12 -0.74
3 0.40 0.14 -0.19
4 0.41 0.31 0.05
5 0.95 0.33 0.65
6 0.98 0.44 0.86
7 1.76 0.76 1.47
8 1.87 1.45 1.53
9 2.24 1.49 2.27
In [198]: n = 5
In [199]: N = (np.partition(df.values.flatten(), n)[:n]).max()
In [200]: (df<=N).idxmin(axis=0)
Out[200]:
blue 1
green 1
red 3
dtype: int64
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
Pretty sure it works with(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though
– MegaBluejay
Nov 15 '18 at 19:55
@MegaBluejay The issue was withnp.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.
– Divakar
Nov 15 '18 at 20:00
add a comment |
Lets say, you are looking at 10 smallest, you can stack and find value_count for the 10 smallest
df.stack().nsmallest(10).index.get_level_values(1).value_counts()
You get
red 5
blue 4
green 1
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
Think you can compare the largest of n-smallest values against the partitioned one and then use idxmin
to leverage the sorted nature -
# Find largest of n smallest numbers
N = (np.partition(df.values.flatten(), n)[:n]).max()
out = (df<=N).idxmin(axis=0)
Sample run -
In [152]: np.random.seed(0)
In [153]: df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2))
for q in ['blue', 'green', 'red'])
In [154]: df
Out[154]:
blue green red
0 -0.98 -0.85 -2.55
1 -0.15 -0.21 -1.45
2 -0.10 0.12 -0.74
3 0.40 0.14 -0.19
4 0.41 0.31 0.05
5 0.95 0.33 0.65
6 0.98 0.44 0.86
7 1.76 0.76 1.47
8 1.87 1.45 1.53
9 2.24 1.49 2.27
In [198]: n = 5
In [199]: N = (np.partition(df.values.flatten(), n)[:n]).max()
In [200]: (df<=N).idxmin(axis=0)
Out[200]:
blue 1
green 1
red 3
dtype: int64
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
Pretty sure it works with(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though
– MegaBluejay
Nov 15 '18 at 19:55
@MegaBluejay The issue was withnp.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.
– Divakar
Nov 15 '18 at 20:00
add a comment |
Think you can compare the largest of n-smallest values against the partitioned one and then use idxmin
to leverage the sorted nature -
# Find largest of n smallest numbers
N = (np.partition(df.values.flatten(), n)[:n]).max()
out = (df<=N).idxmin(axis=0)
Sample run -
In [152]: np.random.seed(0)
In [153]: df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2))
for q in ['blue', 'green', 'red'])
In [154]: df
Out[154]:
blue green red
0 -0.98 -0.85 -2.55
1 -0.15 -0.21 -1.45
2 -0.10 0.12 -0.74
3 0.40 0.14 -0.19
4 0.41 0.31 0.05
5 0.95 0.33 0.65
6 0.98 0.44 0.86
7 1.76 0.76 1.47
8 1.87 1.45 1.53
9 2.24 1.49 2.27
In [198]: n = 5
In [199]: N = (np.partition(df.values.flatten(), n)[:n]).max()
In [200]: (df<=N).idxmin(axis=0)
Out[200]:
blue 1
green 1
red 3
dtype: int64
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
Pretty sure it works with(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though
– MegaBluejay
Nov 15 '18 at 19:55
@MegaBluejay The issue was withnp.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.
– Divakar
Nov 15 '18 at 20:00
add a comment |
Think you can compare the largest of n-smallest values against the partitioned one and then use idxmin
to leverage the sorted nature -
# Find largest of n smallest numbers
N = (np.partition(df.values.flatten(), n)[:n]).max()
out = (df<=N).idxmin(axis=0)
Sample run -
In [152]: np.random.seed(0)
In [153]: df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2))
for q in ['blue', 'green', 'red'])
In [154]: df
Out[154]:
blue green red
0 -0.98 -0.85 -2.55
1 -0.15 -0.21 -1.45
2 -0.10 0.12 -0.74
3 0.40 0.14 -0.19
4 0.41 0.31 0.05
5 0.95 0.33 0.65
6 0.98 0.44 0.86
7 1.76 0.76 1.47
8 1.87 1.45 1.53
9 2.24 1.49 2.27
In [198]: n = 5
In [199]: N = (np.partition(df.values.flatten(), n)[:n]).max()
In [200]: (df<=N).idxmin(axis=0)
Out[200]:
blue 1
green 1
red 3
dtype: int64
Think you can compare the largest of n-smallest values against the partitioned one and then use idxmin
to leverage the sorted nature -
# Find largest of n smallest numbers
N = (np.partition(df.values.flatten(), n)[:n]).max()
out = (df<=N).idxmin(axis=0)
Sample run -
In [152]: np.random.seed(0)
In [153]: df = pd.DataFrame(q: np.sort(np.random.randn(10).round(2))
for q in ['blue', 'green', 'red'])
In [154]: df
Out[154]:
blue green red
0 -0.98 -0.85 -2.55
1 -0.15 -0.21 -1.45
2 -0.10 0.12 -0.74
3 0.40 0.14 -0.19
4 0.41 0.31 0.05
5 0.95 0.33 0.65
6 0.98 0.44 0.86
7 1.76 0.76 1.47
8 1.87 1.45 1.53
9 2.24 1.49 2.27
In [198]: n = 5
In [199]: N = (np.partition(df.values.flatten(), n)[:n]).max()
In [200]: (df<=N).idxmin(axis=0)
Out[200]:
blue 1
green 1
red 3
dtype: int64
edited Nov 15 '18 at 19:59
answered Nov 15 '18 at 19:33
DivakarDivakar
157k1489181
157k1489181
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
Pretty sure it works with(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though
– MegaBluejay
Nov 15 '18 at 19:55
@MegaBluejay The issue was withnp.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.
– Divakar
Nov 15 '18 at 20:00
add a comment |
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
Pretty sure it works with(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though
– MegaBluejay
Nov 15 '18 at 19:55
@MegaBluejay The issue was withnp.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.
– Divakar
Nov 15 '18 at 20:00
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
There appears to be a problem with this: for example with the same data and n=5 its results are 1, 0, 2, while their sum should be 5
– MegaBluejay
Nov 15 '18 at 19:52
Pretty sure it works with
(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though– MegaBluejay
Nov 15 '18 at 19:55
Pretty sure it works with
(df<np.partition(df.values.flatten(), n)[n]).idxmin(axis=0)
though– MegaBluejay
Nov 15 '18 at 19:55
@MegaBluejay The issue was with
np.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.– Divakar
Nov 15 '18 at 20:00
@MegaBluejay The issue was with
np.partition
not giving in sorted order, so we need to sort the smallest partition and then proceed. Fixed. Sort or finding max that is.– Divakar
Nov 15 '18 at 20:00
add a comment |
Lets say, you are looking at 10 smallest, you can stack and find value_count for the 10 smallest
df.stack().nsmallest(10).index.get_level_values(1).value_counts()
You get
red 5
blue 4
green 1
add a comment |
Lets say, you are looking at 10 smallest, you can stack and find value_count for the 10 smallest
df.stack().nsmallest(10).index.get_level_values(1).value_counts()
You get
red 5
blue 4
green 1
add a comment |
Lets say, you are looking at 10 smallest, you can stack and find value_count for the 10 smallest
df.stack().nsmallest(10).index.get_level_values(1).value_counts()
You get
red 5
blue 4
green 1
Lets say, you are looking at 10 smallest, you can stack and find value_count for the 10 smallest
df.stack().nsmallest(10).index.get_level_values(1).value_counts()
You get
red 5
blue 4
green 1
answered Nov 15 '18 at 19:22
VaishaliVaishali
21.7k41336
21.7k41336
add a comment |
add a comment |
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