Filter the rows in a list of tuples using numpy
I am looking for a quicker way to filter out the list of tuples, using numpy and avoiding loops.
A = [(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)]
Now I have to filter it based on the second element, i.e. 4.
If '4' is present form one list 'B', if not form list 'C'.
That is:
B = [(27157, 4),(2857, 4)]
C = [(24814, 0),(1047, 2),(18265, 2),(23854, 2),(36881, 0)]
python arrays numpy indexing
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
asked Nov 12 '18 at 17:37
Gurpreet.S
496
add a comment |
I am looking for a quicker way to filter out the list of tuples, using numpy and avoiding loops.
A = [(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)]
Now I have to filter it based on the second element, i.e. 4.
If '4' is present form one list 'B', if not form list 'C'.
That is:
B = [(27157, 4),(2857, 4)]
C = [(24814, 0),(1047, 2),(18265, 2),(23854, 2),(36881, 0)]
python arrays numpy indexing
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
asked Nov 12 '18 at 17:37
Gurpreet.S
496
If it really is a list (not already an array), a list operation probably will be fastest. There's a significant overhead when creating an array from a list.
– hpaulj
Nov 12 '18 at 20:04
Yes, currently its the list of tuples, on which I am performing the operation using for loop, but I am looking out for a quicker method, so thought of using numpy.
– Gurpreet.S
Nov 12 '18 at 22:00
The kind of thing you are trying to isn't particularly fast, even if you start with an array. But do your own time tests,
– hpaulj
Nov 12 '18 at 22:17
add a comment |
I am looking for a quicker way to filter out the list of tuples, using numpy and avoiding loops.
A = [(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)]
Now I have to filter it based on the second element, i.e. 4.
If '4' is present form one list 'B', if not form list 'C'.
That is:
B = [(27157, 4),(2857, 4)]
C = [(24814, 0),(1047, 2),(18265, 2),(23854, 2),(36881, 0)]
python arrays numpy indexing
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
asked Nov 12 '18 at 17:37
Gurpreet.S
496
I am looking for a quicker way to filter out the list of tuples, using numpy and avoiding loops.
A = [(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)]
Now I have to filter it based on the second element, i.e. 4.
If '4' is present form one list 'B', if not form list 'C'.
That is:
B = [(27157, 4),(2857, 4)]
C = [(24814, 0),(1047, 2),(18265, 2),(23854, 2),(36881, 0)]
python arrays numpy indexing
python arrays numpy indexing
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
asked Nov 12 '18 at 17:37
Gurpreet.S
496
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
asked Nov 12 '18 at 17:37
Gurpreet.S
496
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
edited Nov 12 '18 at 18:13
jpp
91.9k2052102
91.9k2052102
asked Nov 12 '18 at 17:37
Gurpreet.S
496
asked Nov 12 '18 at 17:37
Gurpreet.S
496
asked Nov 12 '18 at 17:37
Gurpreet.S
496
496
If it really is a list (not already an array), a list operation probably will be fastest. There's a significant overhead when creating an array from a list.
– hpaulj
Nov 12 '18 at 20:04
Yes, currently its the list of tuples, on which I am performing the operation using for loop, but I am looking out for a quicker method, so thought of using numpy.
– Gurpreet.S
Nov 12 '18 at 22:00
The kind of thing you are trying to isn't particularly fast, even if you start with an array. But do your own time tests,
– hpaulj
Nov 12 '18 at 22:17
add a comment |
If it really is a list (not already an array), a list operation probably will be fastest. There's a significant overhead when creating an array from a list.
– hpaulj
Nov 12 '18 at 20:04
Yes, currently its the list of tuples, on which I am performing the operation using for loop, but I am looking out for a quicker method, so thought of using numpy.
– Gurpreet.S
Nov 12 '18 at 22:00
The kind of thing you are trying to isn't particularly fast, even if you start with an array. But do your own time tests,
– hpaulj
Nov 12 '18 at 22:17
If it really is a list (not already an array), a list operation probably will be fastest. There's a significant overhead when creating an array from a list.
– hpaulj
Nov 12 '18 at 20:04
If it really is a list (not already an array), a list operation probably will be fastest. There's a significant overhead when creating an array from a list.
– hpaulj
Nov 12 '18 at 20:04
Yes, currently its the list of tuples, on which I am performing the operation using for loop, but I am looking out for a quicker method, so thought of using numpy.
– Gurpreet.S
Nov 12 '18 at 22:00
Yes, currently its the list of tuples, on which I am performing the operation using for loop, but I am looking out for a quicker method, so thought of using numpy.
– Gurpreet.S
Nov 12 '18 at 22:00
The kind of thing you are trying to isn't particularly fast, even if you start with an array. But do your own time tests,
– hpaulj
Nov 12 '18 at 22:17
The kind of thing you are trying to isn't particularly fast, even if you start with an array. But do your own time tests,
– hpaulj
Nov 12 '18 at 22:17
add a comment |
2 Answers
2
active
oldest
votes
With NumPy, you can use Boolean indexing to return arrays:
mask = A[:, 1] == 4
B = A[mask]
C = A[~mask]
This requires your input to be a NumPy array:
A = np.array([(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)])
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
add a comment |
To be fast you must turn you list of tuple in a more efficient data structure. if you want to keep tuples, you can use a structured array :
dt=dtype([('val',int),('key',int)])
B=ndarray(len(A),dt,array(A))
B[B['key']==4] #--> array([(27157, 4), ( 2857, 4)],...
B[B['key']!=4] #--> array([(24814, 0), ( 1047, 2), (18265, 2), (23854, 2), (36881, 0)],...
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
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
With NumPy, you can use Boolean indexing to return arrays:
mask = A[:, 1] == 4
B = A[mask]
C = A[~mask]
This requires your input to be a NumPy array:
A = np.array([(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)])
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
add a comment |
With NumPy, you can use Boolean indexing to return arrays:
mask = A[:, 1] == 4
B = A[mask]
C = A[~mask]
This requires your input to be a NumPy array:
A = np.array([(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)])
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
add a comment |
With NumPy, you can use Boolean indexing to return arrays:
mask = A[:, 1] == 4
B = A[mask]
C = A[~mask]
This requires your input to be a NumPy array:
A = np.array([(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)])
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
With NumPy, you can use Boolean indexing to return arrays:
mask = A[:, 1] == 4
B = A[mask]
C = A[~mask]
This requires your input to be a NumPy array:
A = np.array([(27157, 4),
(24814, 0),
(1047, 2),
(18265, 2),
(2857, 4),
(23854, 2),
(36881, 0)])
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
answered Nov 12 '18 at 17:38
jpp
91.9k2052102
91.9k2052102
add a comment |
add a comment |
To be fast you must turn you list of tuple in a more efficient data structure. if you want to keep tuples, you can use a structured array :
dt=dtype([('val',int),('key',int)])
B=ndarray(len(A),dt,array(A))
B[B['key']==4] #--> array([(27157, 4), ( 2857, 4)],...
B[B['key']!=4] #--> array([(24814, 0), ( 1047, 2), (18265, 2), (23854, 2), (36881, 0)],...
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
add a comment |
To be fast you must turn you list of tuple in a more efficient data structure. if you want to keep tuples, you can use a structured array :
dt=dtype([('val',int),('key',int)])
B=ndarray(len(A),dt,array(A))
B[B['key']==4] #--> array([(27157, 4), ( 2857, 4)],...
B[B['key']!=4] #--> array([(24814, 0), ( 1047, 2), (18265, 2), (23854, 2), (36881, 0)],...
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
add a comment |
To be fast you must turn you list of tuple in a more efficient data structure. if you want to keep tuples, you can use a structured array :
dt=dtype([('val',int),('key',int)])
B=ndarray(len(A),dt,array(A))
B[B['key']==4] #--> array([(27157, 4), ( 2857, 4)],...
B[B['key']!=4] #--> array([(24814, 0), ( 1047, 2), (18265, 2), (23854, 2), (36881, 0)],...
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
To be fast you must turn you list of tuple in a more efficient data structure. if you want to keep tuples, you can use a structured array :
dt=dtype([('val',int),('key',int)])
B=ndarray(len(A),dt,array(A))
B[B['key']==4] #--> array([(27157, 4), ( 2857, 4)],...
B[B['key']!=4] #--> array([(24814, 0), ( 1047, 2), (18265, 2), (23854, 2), (36881, 0)],...
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
answered Nov 12 '18 at 18:11
B. M.
12.9k11934
12.9k11934
add a comment |
add a comment |
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If it really is a list (not already an array), a list operation probably will be fastest. There's a significant overhead when creating an array from a list.
– hpaulj
Nov 12 '18 at 20:04
Yes, currently its the list of tuples, on which I am performing the operation using for loop, but I am looking out for a quicker method, so thought of using numpy.
– Gurpreet.S
Nov 12 '18 at 22:00
The kind of thing you are trying to isn't particularly fast, even if you start with an array. But do your own time tests,
– hpaulj
Nov 12 '18 at 22:17