IndexError: expected dim 0 index in range [1, 3) Error Python
In Python, I am having strange error, where a = [x[1], x[2]] works, but a = x[1:] does not.
>>> out
farray([Y[0], Y[1], Y[2]])
>>> out[1]
Y[1]
>>> remaining_out = [out[1], out[2]]
>>> remaining_out[0]
Y[1]
>>> remaining_out = out[1:]
>>> remaining_out[0]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 485, in __getitem__
nsls = self._norm_slices(fsls) File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 890, in _norm_slices nsls.append(_norm_index(i, fsl, *self.shape[i]))
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 1127, in _norm_index
raise IndexError(fstr.format(dim, start, stop))
IndexError: expected dim 0 index in range [1, 3)
Please help.
python python-3.x
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
add a comment |
In Python, I am having strange error, where a = [x[1], x[2]] works, but a = x[1:] does not.
>>> out
farray([Y[0], Y[1], Y[2]])
>>> out[1]
Y[1]
>>> remaining_out = [out[1], out[2]]
>>> remaining_out[0]
Y[1]
>>> remaining_out = out[1:]
>>> remaining_out[0]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 485, in __getitem__
nsls = self._norm_slices(fsls) File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 890, in _norm_slices nsls.append(_norm_index(i, fsl, *self.shape[i]))
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 1127, in _norm_index
raise IndexError(fstr.format(dim, start, stop))
IndexError: expected dim 0 index in range [1, 3)
Please help.
python python-3.x
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
2
Doesfarrayaccept subscript notations? It does not seem to be a Pythonlisttype.
– BernardL
Nov 26 '18 at 5:40
1
Some support might be found at this Function Array website. That being said, without additional code and a description of your goal, you will not likely have many responses.
– Andy
Nov 26 '18 at 5:41
add a comment |
In Python, I am having strange error, where a = [x[1], x[2]] works, but a = x[1:] does not.
>>> out
farray([Y[0], Y[1], Y[2]])
>>> out[1]
Y[1]
>>> remaining_out = [out[1], out[2]]
>>> remaining_out[0]
Y[1]
>>> remaining_out = out[1:]
>>> remaining_out[0]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 485, in __getitem__
nsls = self._norm_slices(fsls) File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 890, in _norm_slices nsls.append(_norm_index(i, fsl, *self.shape[i]))
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 1127, in _norm_index
raise IndexError(fstr.format(dim, start, stop))
IndexError: expected dim 0 index in range [1, 3)
Please help.
python python-3.x
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
In Python, I am having strange error, where a = [x[1], x[2]] works, but a = x[1:] does not.
>>> out
farray([Y[0], Y[1], Y[2]])
>>> out[1]
Y[1]
>>> remaining_out = [out[1], out[2]]
>>> remaining_out[0]
Y[1]
>>> remaining_out = out[1:]
>>> remaining_out[0]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 485, in __getitem__
nsls = self._norm_slices(fsls) File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 890, in _norm_slices nsls.append(_norm_index(i, fsl, *self.shape[i]))
File "/home/kmshah4/.local/lib/python3.6/site-packages/pyeda/boolalg/bfarray.py", line 1127, in _norm_index
raise IndexError(fstr.format(dim, start, stop))
IndexError: expected dim 0 index in range [1, 3)
Please help.
python python-3.x
python python-3.x
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
asked Nov 26 '18 at 5:29
Karan ShahKaran Shah
1,2361531
1,2361531
2
Doesfarrayaccept subscript notations? It does not seem to be a Pythonlisttype.
– BernardL
Nov 26 '18 at 5:40
1
Some support might be found at this Function Array website. That being said, without additional code and a description of your goal, you will not likely have many responses.
– Andy
Nov 26 '18 at 5:41
add a comment |
2
Doesfarrayaccept subscript notations? It does not seem to be a Pythonlisttype.
– BernardL
Nov 26 '18 at 5:40
1
Some support might be found at this Function Array website. That being said, without additional code and a description of your goal, you will not likely have many responses.
– Andy
Nov 26 '18 at 5:41
2
2
Does
farray accept subscript notations? It does not seem to be a Python list type.– BernardL
Nov 26 '18 at 5:40
Does
farray accept subscript notations? It does not seem to be a Python list type.– BernardL
Nov 26 '18 at 5:40
1
1
Some support might be found at this Function Array website. That being said, without additional code and a description of your goal, you will not likely have many responses.
– Andy
Nov 26 '18 at 5:41
Some support might be found at this Function Array website. That being said, without additional code and a description of your goal, you will not likely have many responses.
– Andy
Nov 26 '18 at 5:41
add a comment |
1 Answer
1
active
oldest
votes
Objects in Python do not automatically support slice notation just because they support indexing. That has to be programmed in. There may be syntax more convenient than what you're using though.
If it's iterable, you can convert to a list first, and then slice it, like
a = [*x][1:]
For especially large iterables (does not seem to be the case here), this may be inefficient. In that case, islice it before you unpack into the list.
from itertools import islice
a = [*islice(x, 1, None)]
But objects do not automatically support iteration just because they support indexing either. If that's the case, you can still iterate manually with range() and a comprehension,
a = [x[i] for i in range(1, len(x))]
Though this isn't really shorter until you have a few more elements to deal with. If you need this pattern a lot you could abstract it into a function.
And finally, objects do not necessarily support len just because they support indexing. If you don't know the length in advance, about all you can do is loop through and catch the LookupError at that point.
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
I did it withlist(x)[1:]. Is that good way?
– Karan Shah
Nov 26 '18 at 7:09
1
It's the same as[*x][1:], pretty much, just a bit more verbose.list(x)would still work on Python 2.7, but the[*x]syntax is newer.
– gilch
Nov 26 '18 at 16:55
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
Objects in Python do not automatically support slice notation just because they support indexing. That has to be programmed in. There may be syntax more convenient than what you're using though.
If it's iterable, you can convert to a list first, and then slice it, like
a = [*x][1:]
For especially large iterables (does not seem to be the case here), this may be inefficient. In that case, islice it before you unpack into the list.
from itertools import islice
a = [*islice(x, 1, None)]
But objects do not automatically support iteration just because they support indexing either. If that's the case, you can still iterate manually with range() and a comprehension,
a = [x[i] for i in range(1, len(x))]
Though this isn't really shorter until you have a few more elements to deal with. If you need this pattern a lot you could abstract it into a function.
And finally, objects do not necessarily support len just because they support indexing. If you don't know the length in advance, about all you can do is loop through and catch the LookupError at that point.
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
I did it withlist(x)[1:]. Is that good way?
– Karan Shah
Nov 26 '18 at 7:09
1
It's the same as[*x][1:], pretty much, just a bit more verbose.list(x)would still work on Python 2.7, but the[*x]syntax is newer.
– gilch
Nov 26 '18 at 16:55
add a comment |
Objects in Python do not automatically support slice notation just because they support indexing. That has to be programmed in. There may be syntax more convenient than what you're using though.
If it's iterable, you can convert to a list first, and then slice it, like
a = [*x][1:]
For especially large iterables (does not seem to be the case here), this may be inefficient. In that case, islice it before you unpack into the list.
from itertools import islice
a = [*islice(x, 1, None)]
But objects do not automatically support iteration just because they support indexing either. If that's the case, you can still iterate manually with range() and a comprehension,
a = [x[i] for i in range(1, len(x))]
Though this isn't really shorter until you have a few more elements to deal with. If you need this pattern a lot you could abstract it into a function.
And finally, objects do not necessarily support len just because they support indexing. If you don't know the length in advance, about all you can do is loop through and catch the LookupError at that point.
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
I did it withlist(x)[1:]. Is that good way?
– Karan Shah
Nov 26 '18 at 7:09
1
It's the same as[*x][1:], pretty much, just a bit more verbose.list(x)would still work on Python 2.7, but the[*x]syntax is newer.
– gilch
Nov 26 '18 at 16:55
add a comment |
Objects in Python do not automatically support slice notation just because they support indexing. That has to be programmed in. There may be syntax more convenient than what you're using though.
If it's iterable, you can convert to a list first, and then slice it, like
a = [*x][1:]
For especially large iterables (does not seem to be the case here), this may be inefficient. In that case, islice it before you unpack into the list.
from itertools import islice
a = [*islice(x, 1, None)]
But objects do not automatically support iteration just because they support indexing either. If that's the case, you can still iterate manually with range() and a comprehension,
a = [x[i] for i in range(1, len(x))]
Though this isn't really shorter until you have a few more elements to deal with. If you need this pattern a lot you could abstract it into a function.
And finally, objects do not necessarily support len just because they support indexing. If you don't know the length in advance, about all you can do is loop through and catch the LookupError at that point.
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
Objects in Python do not automatically support slice notation just because they support indexing. That has to be programmed in. There may be syntax more convenient than what you're using though.
If it's iterable, you can convert to a list first, and then slice it, like
a = [*x][1:]
For especially large iterables (does not seem to be the case here), this may be inefficient. In that case, islice it before you unpack into the list.
from itertools import islice
a = [*islice(x, 1, None)]
But objects do not automatically support iteration just because they support indexing either. If that's the case, you can still iterate manually with range() and a comprehension,
a = [x[i] for i in range(1, len(x))]
Though this isn't really shorter until you have a few more elements to deal with. If you need this pattern a lot you could abstract it into a function.
And finally, objects do not necessarily support len just because they support indexing. If you don't know the length in advance, about all you can do is loop through and catch the LookupError at that point.
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
edited Nov 26 '18 at 17:02
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
answered Nov 26 '18 at 6:22
gilchgilch
3,8901716
3,8901716
I did it withlist(x)[1:]. Is that good way?
– Karan Shah
Nov 26 '18 at 7:09
1
It's the same as[*x][1:], pretty much, just a bit more verbose.list(x)would still work on Python 2.7, but the[*x]syntax is newer.
– gilch
Nov 26 '18 at 16:55
add a comment |
I did it withlist(x)[1:]. Is that good way?
– Karan Shah
Nov 26 '18 at 7:09
1
It's the same as[*x][1:], pretty much, just a bit more verbose.list(x)would still work on Python 2.7, but the[*x]syntax is newer.
– gilch
Nov 26 '18 at 16:55
I did it with
list(x)[1:]. Is that good way?– Karan Shah
Nov 26 '18 at 7:09
I did it with
list(x)[1:]. Is that good way?– Karan Shah
Nov 26 '18 at 7:09
1
1
It's the same as
[*x][1:], pretty much, just a bit more verbose. list(x) would still work on Python 2.7, but the [*x] syntax is newer.– gilch
Nov 26 '18 at 16:55
It's the same as
[*x][1:], pretty much, just a bit more verbose. list(x) would still work on Python 2.7, but the [*x] syntax is newer.– gilch
Nov 26 '18 at 16:55
add a comment |
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2
Does
farrayaccept subscript notations? It does not seem to be a Pythonlisttype.– BernardL
Nov 26 '18 at 5:40
1
Some support might be found at this Function Array website. That being said, without additional code and a description of your goal, you will not likely have many responses.
– Andy
Nov 26 '18 at 5:41