本文整理汇总了Python中pandas.core.common._not_none方法的典型用法代码示例。如果您正苦于以下问题：Python common._not_none方法的具体用法？Python common._not_none怎么用？Python common._not_none使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类pandas.core.common的用法示例。
在下文中一共展示了common._not_none方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: _concat_objects
# 需要导入模块: from pandas.core import common [as 别名]
# 或者: from pandas.core.common import _not_none [as 别名]
def _concat_objects(self, keys, values, not_indexed_same=False):
from pandas.core.reshape.concat import concat
def reset_identity(values):
# reset the identities of the components
# of the values to prevent aliasing
for v in com._not_none(*values):
ax = v._get_axis(self.axis)
ax._reset_identity()
return values
if not not_indexed_same:
result = concat(values, axis=self.axis)
ax = self._selected_obj._get_axis(self.axis)
if isinstance(result, Series):
result = result.reindex(ax)
else:
# this is a very unfortunate situation
# we have a multi-index that is NOT lexsorted
# and we have a result which is duplicated
# we can't reindex, so we resort to this
# GH 14776
if isinstance(ax, MultiIndex) and not ax.is_unique:
indexer = algorithms.unique1d(
result.index.get_indexer_for(ax.values))
result = result.take(indexer, axis=self.axis)
else:
result = result.reindex(ax, axis=self.axis)
elif self.group_keys:
values = reset_identity(values)
if self.as_index:
# possible MI return case
group_keys = keys
group_levels = self.grouper.levels
group_names = self.grouper.names
result = concat(values, axis=self.axis, keys=group_keys,
levels=group_levels, names=group_names,
sort=False)
else:
# GH5610, returns a MI, with the first level being a
# range index
keys = list(range(len(values)))
result = concat(values, axis=self.axis, keys=keys)
else:
values = reset_identity(values)
result = concat(values, axis=self.axis)
if (isinstance(result, Series) and
getattr(self, '_selection_name', None) is not None):
result.name = self._selection_name
return result 示例2: _join_multi
# 需要导入模块: from pandas.core import common [as 别名]
# 或者: from pandas.core.common import _not_none [as 别名]
def _join_multi(self, other, how, return_indexers=True):
from .multi import MultiIndex
self_is_mi = isinstance(self, MultiIndex)
other_is_mi = isinstance(other, MultiIndex)
# figure out join names
self_names = com._not_none(*self.names)
other_names = com._not_none(*other.names)
overlap = list(set(self_names) & set(other_names))
# need at least 1 in common, but not more than 1
if not len(overlap):
raise ValueError("cannot join with no level specified and no "
"overlapping names")
if len(overlap) > 1:
raise NotImplementedError("merging with more than one level "
"overlap on a multi-index is not "
"implemented")
jl = overlap[0]
# make the indices into mi's that match
if not (self_is_mi and other_is_mi):
flip_order = False
if self_is_mi:
self, other = other, self
flip_order = True
# flip if join method is right or left
how = {'right': 'left', 'left': 'right'}.get(how, how)
level = other.names.index(jl)
result = self._join_level(other, level, how=how,
return_indexers=return_indexers)
if flip_order:
if isinstance(result, tuple):
return result[0], result[2], result[1]
return result
# 2 multi-indexes
raise NotImplementedError("merging with both multi-indexes is not "
"implemented")