Mixing
Check and cast multiindex types
How can I check a Pandas' Multiindex type for each of its levels?
I'm trying to access a specific level in order to check whether its type is what I want it to be, and if not, cast it to an int type.
I've tried df.index.info() with no success. I've also checked the methods and attributes described in the API Reference, though I don't see any mention to it.
python pandas
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
add a comment |
How can I check a Pandas' Multiindex type for each of its levels?
I'm trying to access a specific level in order to check whether its type is what I want it to be, and if not, cast it to an int type.
I've tried df.index.info() with no success. I've also checked the methods and attributes described in the API Reference, though I don't see any mention to it.
python pandas
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
add a comment |
How can I check a Pandas' Multiindex type for each of its levels?
I'm trying to access a specific level in order to check whether its type is what I want it to be, and if not, cast it to an int type.
I've tried df.index.info() with no success. I've also checked the methods and attributes described in the API Reference, though I don't see any mention to it.
python pandas
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
How can I check a Pandas' Multiindex type for each of its levels?
I'm trying to access a specific level in order to check whether its type is what I want it to be, and if not, cast it to an int type.
I've tried df.index.info() with no success. I've also checked the methods and attributes described in the API Reference, though I don't see any mention to it.
python pandas
python pandas
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
asked Nov 21 '18 at 14:31
jimijazzjimijazz
954822
954822
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Setup
idx = pd.MultiIndex.from_product([range(2), [*'XY']], names=['lvl0', 'lvl1'])
df = pd.DataFrame(1, idx, idx)
df
lvl0 0 1
lvl1 X Y X Y
lvl0 lvl1
0 X 1 1 1 1
Y 1 1 1 1
1 X 1 1 1 1
Y 1 1 1 1
Basic Anatomy of a MultiIndex
levels
Is a frozenlist of pandas.Index objects. Each of these pandas.Index objects should contain unique values. If these level index objects are not unique, something is probably broken.
[*map(type, df.index.levels)]
[pandas.core.indexes.numeric.Int64Index, pandas.core.indexes.base.Index]
you can get at their dtype
[l.dtype for l in df.index.levels]
[dtype('int64'), dtype('O')]
labels
This is a frozenlist of arrays. There is a one label array for each level index. The corresponding label array contains reference to which level values are being displayed.
[*map(type, df.index.labels)]
[pandas.core.indexes.frozen.FrozenNDArray,
pandas.core.indexes.frozen.FrozenNDArray]
print(*df.index.labels, sep='n')
FrozenNDArray([0, 0, 1, 1], dtype='int8')
FrozenNDArray([0, 1, 0, 1], dtype='int8')
get_level_values
You can access the values in an index with get_level_values
df.index.get_level_values(1)
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
Which would be the same as slicing the level object with the label object
df.index.levels[1][df.index.labels[1]]
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
1
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
add a comment |
Using get_level_values
df.index.get_level_values(0).dtype
Out[19]: dtype('int64')
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
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
Setup
idx = pd.MultiIndex.from_product([range(2), [*'XY']], names=['lvl0', 'lvl1'])
df = pd.DataFrame(1, idx, idx)
df
lvl0 0 1
lvl1 X Y X Y
lvl0 lvl1
0 X 1 1 1 1
Y 1 1 1 1
1 X 1 1 1 1
Y 1 1 1 1
Basic Anatomy of a MultiIndex
levels
Is a frozenlist of pandas.Index objects. Each of these pandas.Index objects should contain unique values. If these level index objects are not unique, something is probably broken.
[*map(type, df.index.levels)]
[pandas.core.indexes.numeric.Int64Index, pandas.core.indexes.base.Index]
you can get at their dtype
[l.dtype for l in df.index.levels]
[dtype('int64'), dtype('O')]
labels
This is a frozenlist of arrays. There is a one label array for each level index. The corresponding label array contains reference to which level values are being displayed.
[*map(type, df.index.labels)]
[pandas.core.indexes.frozen.FrozenNDArray,
pandas.core.indexes.frozen.FrozenNDArray]
print(*df.index.labels, sep='n')
FrozenNDArray([0, 0, 1, 1], dtype='int8')
FrozenNDArray([0, 1, 0, 1], dtype='int8')
get_level_values
You can access the values in an index with get_level_values
df.index.get_level_values(1)
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
Which would be the same as slicing the level object with the label object
df.index.levels[1][df.index.labels[1]]
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
1
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
add a comment |
Setup
idx = pd.MultiIndex.from_product([range(2), [*'XY']], names=['lvl0', 'lvl1'])
df = pd.DataFrame(1, idx, idx)
df
lvl0 0 1
lvl1 X Y X Y
lvl0 lvl1
0 X 1 1 1 1
Y 1 1 1 1
1 X 1 1 1 1
Y 1 1 1 1
Basic Anatomy of a MultiIndex
levels
Is a frozenlist of pandas.Index objects. Each of these pandas.Index objects should contain unique values. If these level index objects are not unique, something is probably broken.
[*map(type, df.index.levels)]
[pandas.core.indexes.numeric.Int64Index, pandas.core.indexes.base.Index]
you can get at their dtype
[l.dtype for l in df.index.levels]
[dtype('int64'), dtype('O')]
labels
This is a frozenlist of arrays. There is a one label array for each level index. The corresponding label array contains reference to which level values are being displayed.
[*map(type, df.index.labels)]
[pandas.core.indexes.frozen.FrozenNDArray,
pandas.core.indexes.frozen.FrozenNDArray]
print(*df.index.labels, sep='n')
FrozenNDArray([0, 0, 1, 1], dtype='int8')
FrozenNDArray([0, 1, 0, 1], dtype='int8')
get_level_values
You can access the values in an index with get_level_values
df.index.get_level_values(1)
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
Which would be the same as slicing the level object with the label object
df.index.levels[1][df.index.labels[1]]
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
1
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
add a comment |
Setup
idx = pd.MultiIndex.from_product([range(2), [*'XY']], names=['lvl0', 'lvl1'])
df = pd.DataFrame(1, idx, idx)
df
lvl0 0 1
lvl1 X Y X Y
lvl0 lvl1
0 X 1 1 1 1
Y 1 1 1 1
1 X 1 1 1 1
Y 1 1 1 1
Basic Anatomy of a MultiIndex
levels
Is a frozenlist of pandas.Index objects. Each of these pandas.Index objects should contain unique values. If these level index objects are not unique, something is probably broken.
[*map(type, df.index.levels)]
[pandas.core.indexes.numeric.Int64Index, pandas.core.indexes.base.Index]
you can get at their dtype
[l.dtype for l in df.index.levels]
[dtype('int64'), dtype('O')]
labels
This is a frozenlist of arrays. There is a one label array for each level index. The corresponding label array contains reference to which level values are being displayed.
[*map(type, df.index.labels)]
[pandas.core.indexes.frozen.FrozenNDArray,
pandas.core.indexes.frozen.FrozenNDArray]
print(*df.index.labels, sep='n')
FrozenNDArray([0, 0, 1, 1], dtype='int8')
FrozenNDArray([0, 1, 0, 1], dtype='int8')
get_level_values
You can access the values in an index with get_level_values
df.index.get_level_values(1)
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
Which would be the same as slicing the level object with the label object
df.index.levels[1][df.index.labels[1]]
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
Setup
idx = pd.MultiIndex.from_product([range(2), [*'XY']], names=['lvl0', 'lvl1'])
df = pd.DataFrame(1, idx, idx)
df
lvl0 0 1
lvl1 X Y X Y
lvl0 lvl1
0 X 1 1 1 1
Y 1 1 1 1
1 X 1 1 1 1
Y 1 1 1 1
Basic Anatomy of a MultiIndex
levels
Is a frozenlist of pandas.Index objects. Each of these pandas.Index objects should contain unique values. If these level index objects are not unique, something is probably broken.
[*map(type, df.index.levels)]
[pandas.core.indexes.numeric.Int64Index, pandas.core.indexes.base.Index]
you can get at their dtype
[l.dtype for l in df.index.levels]
[dtype('int64'), dtype('O')]
labels
This is a frozenlist of arrays. There is a one label array for each level index. The corresponding label array contains reference to which level values are being displayed.
[*map(type, df.index.labels)]
[pandas.core.indexes.frozen.FrozenNDArray,
pandas.core.indexes.frozen.FrozenNDArray]
print(*df.index.labels, sep='n')
FrozenNDArray([0, 0, 1, 1], dtype='int8')
FrozenNDArray([0, 1, 0, 1], dtype='int8')
get_level_values
You can access the values in an index with get_level_values
df.index.get_level_values(1)
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
Which would be the same as slicing the level object with the label object
df.index.levels[1][df.index.labels[1]]
Index(['X', 'Y', 'X', 'Y'], dtype='object', name='lvl1')
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
answered Nov 21 '18 at 14:52
piRSquaredpiRSquared
155k22150294
155k22150294
1
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
add a comment |
1
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
1
1
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
happy thanksgiving:-)
– Wen-Ben
Nov 21 '18 at 14:57
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
You too! And thanks (-:
– piRSquared
Nov 21 '18 at 14:58
add a comment |
Using get_level_values
df.index.get_level_values(0).dtype
Out[19]: dtype('int64')
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
add a comment |
Using get_level_values
df.index.get_level_values(0).dtype
Out[19]: dtype('int64')
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
add a comment |
Using get_level_values
df.index.get_level_values(0).dtype
Out[19]: dtype('int64')
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
Using get_level_values
df.index.get_level_values(0).dtype
Out[19]: dtype('int64')
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
answered Nov 21 '18 at 14:49
Wen-BenWen-Ben
110k83266
110k83266
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
add a comment |
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
this is a good answer, but for the sake of completeness I've marked the other one as accepted. Thanks anyways!
– jimijazz
Nov 22 '18 at 13:18
add a comment |
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