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Skip idx.is_lexsorted() when pandas version is larger than 1.3.0. (#973)

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* Skip idx.is_lexsorted() when pandas version is larger than 1.3.0. The future warning is annoying.

* Skip idx.is_lexsorted() when pandas version is larger than 1.3.0. The future warning is annoying.

* Rewrite code.
This commit is contained in:
Chia-hung Tai
2022-03-13 23:24:54 +08:00
committed by GitHub
parent 829ad9f5e9
commit 776b0c5bb4
1 changed files with 11 additions and 4 deletions
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15
qlib/utils/__init__.py
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@@ -32,11 +32,14 @@ from pathlib import Path
from typing import List, Dict, Union, Tuple, Any, Text, Optional, Callable
from types import ModuleType
from urllib.parse import urlparse
from packaging import version
from .file import get_or_create_path, save_multiple_parts_file, unpack_archive_with_buffer, get_tmp_file_with_buffer
from ..config import C
from ..log import get_module_logger, set_log_with_config
log = get_module_logger("utils")
# MultiIndex.is_lexsorted() is a deprecated method in Pandas 1.3.0.
is_deprecated_lexsorted_pandas = version.parse(pd.__version__) > version.parse("1.3.0")
#################### Server ####################
@@ -788,11 +791,15 @@ def lazy_sort_index(df: pd.DataFrame, axis=0) -> pd.DataFrame:
sorted dataframe
"""
idx = df.index if axis == 0 else df.columns
# NOTE: MultiIndex.is_lexsorted() is a deprecated method in Pandas 1.3.0 and is suggested to be replaced by MultiIndex.is_monotonic_increasing (see discussion here: https://github.com/pandas-dev/pandas/issues/32259). However, in case older versions of Pandas is implemented, MultiIndex.is_lexsorted() is necessary to prevent certain fatal errors.
if idx.is_monotonic_increasing and not (isinstance(idx, pd.MultiIndex) and not idx.is_lexsorted()):
return df
else:
if (
not idx.is_monotonic_increasing
or not is_deprecated_lexsorted_pandas
and isinstance(idx, pd.MultiIndex)
and not idx.is_lexsorted()
): # this case is for the old version
return df.sort_index(axis=axis)
else:
return df
FLATTEN_TUPLE = "_FLATTEN_TUPLE"