Is there a build in function in Python to get the size of a list like C++?
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When a list has only 1 dimension, len() could be used to get the size of it, is there a function to get the size of a nested list?
for example, test_list = [[1], [2], [3]], len(test_list) return 3, how to get (3, 1) ?
python
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
add a comment |
When a list has only 1 dimension, len() could be used to get the size of it, is there a function to get the size of a nested list?
for example, test_list = [[1], [2], [3]], len(test_list) return 3, how to get (3, 1) ?
python
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
6
Why should the result be(3, 1)? The sublists can all be different lengths, and can be nested to any depth. They can even be recursive.
– PM 2Ring
Nov 23 '18 at 15:34
1
You could usenumpy:np.array([[1], [2], [3]]).shape == (3, 1). If the inner lists don't have a consistent length, this will create an array of objects, but if they're consistent you'll get a multi-dimensional array.
– jonrsharpe
Nov 23 '18 at 15:35
list(map(len, values))=>[1, 1, 1]
– Peter Wood
Nov 23 '18 at 15:54
@PM 2Ring, eh, I assumed the sublists are in same length, but not realized to figure it out, it is my bad, Thank you for your kind reminder.
– buxizhizhoum
Nov 24 '18 at 1:33
add a comment |
When a list has only 1 dimension, len() could be used to get the size of it, is there a function to get the size of a nested list?
for example, test_list = [[1], [2], [3]], len(test_list) return 3, how to get (3, 1) ?
python
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
When a list has only 1 dimension, len() could be used to get the size of it, is there a function to get the size of a nested list?
for example, test_list = [[1], [2], [3]], len(test_list) return 3, how to get (3, 1) ?
python
python
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
edited Nov 24 '18 at 1:35
buxizhizhoum
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
asked Nov 23 '18 at 15:32
buxizhizhoumbuxizhizhoum
373512
373512
6
Why should the result be(3, 1)? The sublists can all be different lengths, and can be nested to any depth. They can even be recursive.
– PM 2Ring
Nov 23 '18 at 15:34
1
You could usenumpy:np.array([[1], [2], [3]]).shape == (3, 1). If the inner lists don't have a consistent length, this will create an array of objects, but if they're consistent you'll get a multi-dimensional array.
– jonrsharpe
Nov 23 '18 at 15:35
list(map(len, values))=>[1, 1, 1]
– Peter Wood
Nov 23 '18 at 15:54
@PM 2Ring, eh, I assumed the sublists are in same length, but not realized to figure it out, it is my bad, Thank you for your kind reminder.
– buxizhizhoum
Nov 24 '18 at 1:33
add a comment |
6
Why should the result be(3, 1)? The sublists can all be different lengths, and can be nested to any depth. They can even be recursive.
– PM 2Ring
Nov 23 '18 at 15:34
1
You could usenumpy:np.array([[1], [2], [3]]).shape == (3, 1). If the inner lists don't have a consistent length, this will create an array of objects, but if they're consistent you'll get a multi-dimensional array.
– jonrsharpe
Nov 23 '18 at 15:35
list(map(len, values))=>[1, 1, 1]
– Peter Wood
Nov 23 '18 at 15:54
@PM 2Ring, eh, I assumed the sublists are in same length, but not realized to figure it out, it is my bad, Thank you for your kind reminder.
– buxizhizhoum
Nov 24 '18 at 1:33
6
6
Why should the result be
(3, 1)? The sublists can all be different lengths, and can be nested to any depth. They can even be recursive.– PM 2Ring
Nov 23 '18 at 15:34
Why should the result be
(3, 1)? The sublists can all be different lengths, and can be nested to any depth. They can even be recursive.– PM 2Ring
Nov 23 '18 at 15:34
1
1
You could use
numpy: np.array([[1], [2], [3]]).shape == (3, 1). If the inner lists don't have a consistent length, this will create an array of objects, but if they're consistent you'll get a multi-dimensional array.– jonrsharpe
Nov 23 '18 at 15:35
You could use
numpy: np.array([[1], [2], [3]]).shape == (3, 1). If the inner lists don't have a consistent length, this will create an array of objects, but if they're consistent you'll get a multi-dimensional array.– jonrsharpe
Nov 23 '18 at 15:35
list(map(len, values)) => [1, 1, 1]– Peter Wood
Nov 23 '18 at 15:54
list(map(len, values)) => [1, 1, 1]– Peter Wood
Nov 23 '18 at 15:54
@PM 2Ring, eh, I assumed the sublists are in same length, but not realized to figure it out, it is my bad, Thank you for your kind reminder.
– buxizhizhoum
Nov 24 '18 at 1:33
@PM 2Ring, eh, I assumed the sublists are in same length, but not realized to figure it out, it is my bad, Thank you for your kind reminder.
– buxizhizhoum
Nov 24 '18 at 1:33
add a comment |
2 Answers
2
active
oldest
votes
It's because amount of elements of your array is indeed 3, if you'd like to check depths of its contents you could do (note that depth of second dimension will be maximum length of its contents)
test_list = [[1], [2], [3]]
print((len(test_list), max(len(i) for i in test_list))) # -> (3, 1)
Or using numpy
import numpy as np
test_list = np.array([[1], [2], [3]])
print(test_list.shape) # -> (3, 1)
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
add a comment |
If you use a numpy array instead of list then you can get the shape which will be (3, 1).
import numpy as np
a = np.array([[1], [2], [3]])
print(a.shape)
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
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
It's because amount of elements of your array is indeed 3, if you'd like to check depths of its contents you could do (note that depth of second dimension will be maximum length of its contents)
test_list = [[1], [2], [3]]
print((len(test_list), max(len(i) for i in test_list))) # -> (3, 1)
Or using numpy
import numpy as np
test_list = np.array([[1], [2], [3]])
print(test_list.shape) # -> (3, 1)
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
add a comment |
It's because amount of elements of your array is indeed 3, if you'd like to check depths of its contents you could do (note that depth of second dimension will be maximum length of its contents)
test_list = [[1], [2], [3]]
print((len(test_list), max(len(i) for i in test_list))) # -> (3, 1)
Or using numpy
import numpy as np
test_list = np.array([[1], [2], [3]])
print(test_list.shape) # -> (3, 1)
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
add a comment |
It's because amount of elements of your array is indeed 3, if you'd like to check depths of its contents you could do (note that depth of second dimension will be maximum length of its contents)
test_list = [[1], [2], [3]]
print((len(test_list), max(len(i) for i in test_list))) # -> (3, 1)
Or using numpy
import numpy as np
test_list = np.array([[1], [2], [3]])
print(test_list.shape) # -> (3, 1)
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
It's because amount of elements of your array is indeed 3, if you'd like to check depths of its contents you could do (note that depth of second dimension will be maximum length of its contents)
test_list = [[1], [2], [3]]
print((len(test_list), max(len(i) for i in test_list))) # -> (3, 1)
Or using numpy
import numpy as np
test_list = np.array([[1], [2], [3]])
print(test_list.shape) # -> (3, 1)
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
answered Nov 23 '18 at 15:37
Filip MłynarskiFilip Młynarski
2,0241415
2,0241415
add a comment |
add a comment |
If you use a numpy array instead of list then you can get the shape which will be (3, 1).
import numpy as np
a = np.array([[1], [2], [3]])
print(a.shape)
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
add a comment |
If you use a numpy array instead of list then you can get the shape which will be (3, 1).
import numpy as np
a = np.array([[1], [2], [3]])
print(a.shape)
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
add a comment |
If you use a numpy array instead of list then you can get the shape which will be (3, 1).
import numpy as np
a = np.array([[1], [2], [3]])
print(a.shape)
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
If you use a numpy array instead of list then you can get the shape which will be (3, 1).
import numpy as np
a = np.array([[1], [2], [3]])
print(a.shape)
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
edited Nov 23 '18 at 15:43
Jean-François Fabre♦
106k1058116
106k1058116
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
answered Nov 23 '18 at 15:37
Attila BognárAttila Bognár
1299
1299
add a comment |
add a comment |
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6
Why should the result be
(3, 1)? The sublists can all be different lengths, and can be nested to any depth. They can even be recursive.– PM 2Ring
Nov 23 '18 at 15:34
1
You could use
numpy:np.array([[1], [2], [3]]).shape == (3, 1). If the inner lists don't have a consistent length, this will create an array of objects, but if they're consistent you'll get a multi-dimensional array.– jonrsharpe
Nov 23 '18 at 15:35
list(map(len, values))=>[1, 1, 1]– Peter Wood
Nov 23 '18 at 15:54
@PM 2Ring, eh, I assumed the sublists are in same length, but not realized to figure it out, it is my bad, Thank you for your kind reminder.
– buxizhizhoum
Nov 24 '18 at 1:33