gpt4 book ai didi

Python:规范化 pandas DataFrame 的一些列

转载 作者:太空狗 更新时间:2023-10-30 03:02:28 24 4
gpt4 key购买 nike

我有一个 DataFrame,我想使用另一个任意列从中规范化一些任意列:

import itertools as it
import numpy as np
import pandas as pd

header = tuple(['h_seqNum', 'h_stamp', 'user_id'])
joints = tuple(['head', 'neck', 'torso'])
attribs = tuple(['pos_x','pos_y','pos_z'])

all_columns = it.izip(*it.product(joints, attribs))
multiind_first = list(it.chain(['header']*len(header), all_columns.next(), ['pose',]))
multiind_second = list(it.chain(header, all_columns.next(), ['pose',]))

df = pd.DataFrame(np.random.rand(65).reshape(5,13), columns = pd.MultiIndex.from_arrays([multiind_first, multiind_second], names=['joint', 'attrib']))

生成的 DataFrame 是这样的:

joint    header                            head                       neck                       torso                      pose
attrib h_seqNum h_stamp user_id pos_x pos_y pos_z pos_x pos_y pos_z pos_x pos_y pos_z pose
0 0.681 0.059 0.607 0.093 0.504 0.975 0.317 0.739 0.129 0.759 0.254 0.814 1
1 0.914 0.420 0.305 0.242 0.700 0.180 0.324 0.171 0.477 0.943 0.877 0.069 0
2 0.522 0.395 0.118 0.739 0.653 0.326 0.947 0.517 0.036 0.647 0.079 0.227 0
3 0.475 0.815 0.792 0.208 0.472 0.427 0.213 0.544 0.440 0.033 0.636 0.527 2
4 0.767 0.774 0.983 0.646 0.949 0.947 0.402 0.015 0.913 0.734 0.192 0.032 0

我想使用另一个任意关节(例如“躯干”)归一化属于任意关节(例如“头部”)的所有列(属性)。例如类似的东西。

df['head'] = df['head'] - df['torso']
df['neck'] = df['neck'] - df['torso']
# Note that torso remains "unnormalized"

为此我写了一个函数:

def normalize_joints(df, from_joint):
joint_names = set(joints) - set([from_joint,])
for j in list(joint_names):
df[j] = df[j] - df[norm_name]

但是,当我执行此函数时,出现以下错误:

normalize_joints(df, 'torso')

---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-414-47f39f04716d> in <module>()
----> 1 normalize_joints(df, 'torso')

<ipython-input-407-cf13a67fabd8> in normalize_joints(df, from_joint)
2 joint_names = set(joints) - set([from_joint,])
3 for j in list(joint_names):
----> 4 df[j] = df[j] - df[from_joint]

/Library/Python/2.7/site-packages/pandas/core/frame.pyc in __setitem__(self, key, value)
2117 fill_value, limit, takeable=takeable)
2118
-> 2119 return frame
2120
2121 def _reindex_index(self, new_index, method, copy, level, fill_value=NA,

/Library/Python/2.7/site-packages/pandas/core/frame.pyc in _set_item(self, key, value)
2164 @Appender(_shared_docs['reindex_axis'] % _shared_doc_kwargs)
2165 def reindex_axis(self, labels, axis=0, method=None, level=None, copy=True,
-> 2166 limit=None, fill_value=np.nan):
2167 return super(DataFrame, self).reindex_axis(labels=labels, axis=axis,
2168 method=method, level=level,

/Library/Python/2.7/site-packages/pandas/core/generic.pyc in _set_item(self, key, value)
677
678 __bool__ = __nonzero__
--> 679
680 def bool(self):
681 """ Return the bool of a single element PandasObject

/Library/Python/2.7/site-packages/pandas/core/internals.pyc in set(self, item, value)
1768 def sp_index(self):
1769 return self.values.sp_index
-> 1770
1771 @property
1772 def kind(self):

/Library/Python/2.7/site-packages/pandas/core/internals.pyc in _reset_ref_locs(self)
1054 # see if we can align other
1055 if hasattr(other, 'reindex_axis'):
-> 1056 if align:
1057 axis = getattr(other, '_info_axis_number', 0)
1058 other = other.reindex_axis(self.items, axis=axis,

/Library/Python/2.7/site-packages/pandas/core/internals.pyc in _rebuild_ref_locs(self)
1062
1063 # make sure that we can broadcast
-> 1064 is_transposed = False
1065 if hasattr(other, 'ndim') and hasattr(values, 'ndim'):
1066 if values.ndim != other.ndim or values.shape == other.shape[::-1]:

AttributeError: _ref_locs

经过多次尝试,我无法找到错误的根源。如果我执行操作

df['head'] - df['torso']

它返回一个具有正确结果的 DataFrame。但是,当我尝试将此 DataFrame 分配给 df['head'] 时,出现了之前显示的错误。

有什么办法可以完成这个作业吗?

此外,我想知道是否有比我正在尝试的方法更好的方法来执行相同的规范化。也许使用 groupby,然后将规范化函数应用于选定的 DataFrame?

编辑:

这个错误发生在 numpy 1.6 和 pandas 0.12

升级到 numpy 1.8 和 pandas 0.13 后,以下操作有效:

df['head'] = df['head'] - df['torso']

最佳答案

问题是您的列是 MultiIndex 的实例,试试这个:

def normalize_joints(df, from_joint):
joint_names = set(joints) - set([from_joint,])
for j in list(joint_names):
keys = [(j,c) for c in attribs]
df[keys] = df[j] - df[from_joint]

print df
normalize_joints(df, 'torso')
print df

输出:

joint     header                          head                          neck                         torso                          pose
attrib h_seqNum h_stamp user_id pos_x pos_y pos_z pos_x pos_y pos_z pos_x pos_y pos_z pose
0 0.067366 0.957394 0.983969 0.602662 0.505270 0.990675 0.753841 0.598397 0.846479 0.757155 0.220009 0.328470 0.686525
1 0.806405 0.800388 0.302178 0.935559 0.180360 0.322767 0.230457 0.617555 0.602589 0.109482 0.181803 0.311266 0.929481
2 0.649677 0.237286 0.963088 0.370463 0.471590 0.489256 0.060383 0.070885 0.858312 0.306232 0.511731 0.257015 0.283287
3 0.054800 0.127925 0.099985 0.700160 0.211256 0.026782 0.820380 0.922593 0.600130 0.100745 0.418157 0.869735 0.597275
4 0.678372 0.334520 0.247894 0.616133 0.914610 0.229628 0.317488 0.224910 0.620222 0.952499 0.946568 0.539502 0.838473
joint header head neck torso pose
attrib h_seqNum h_stamp user_id pos_x pos_y pos_z pos_x pos_y pos_z pos_x pos_y pos_z pose
0 0.067366 0.957394 0.983969 -0.154493 0.285261 0.662205 -0.003314 0.378387 0.518009 0.757155 0.220009 0.328470 0.686525
1 0.806405 0.800388 0.302178 0.826077 -0.001443 0.011501 0.120975 0.435752 0.291322 0.109482 0.181803 0.311266 0.929481
2 0.649677 0.237286 0.963088 0.064231 -0.040141 0.232241 -0.245850 -0.440846 0.601297 0.306232 0.511731 0.257015 0.283287
3 0.054800 0.127925 0.099985 0.599414 -0.206900 -0.842953 0.719635 0.504436 -0.269605 0.100745 0.418157 0.869735 0.597275
4 0.678372 0.334520 0.247894 -0.336366 -0.031958 -0.309874 -0.635011 -0.721658 0.080719 0.952499 0.946568 0.539502 0.838473

关于Python:规范化 pandas DataFrame 的一些列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21832169/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com