如何在包含空列表的列表上垂直平均？
How to average vertically over a list containing a void list?
我有一个包含以下内容的列表：
list1 = [(4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)]
我需要这些值在垂直轴上的平均值，这样输出看起来像：
[(average_col1, average_col2, average_col3, average_col4, average_col5, average_col6)]
然而，np.mean(list1, axis=1)命令returns:
IndexError: tuple index out of range
因此我尝试使用以下方法创建一个 numpy 数组：
a = np.array(list1)
a = array([ (4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)],
dtype=[('col1', '<f8'), ('col2', '<f8'), ('col3', '<f8'), ('col4', '<f8'), ('col5', '<i4'), ('col6', '<i4')])
如果我使用与上面相同的平均命令 returns:
IndexError: tuple index out of range
因此我不确定从这里开始做什么。
您可以在不使用 numpy 的情况下为您的第一个列表尝试此操作：
averages = [sum(i)/float(len(i)) for i in zip(*list)]
您在尝试调用 np.mean 时使用了括号 ([]) 而不是圆括号 (())。这段代码应该做你想做的事：
import numpy as np
list1 = [(4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)]
means = np.mean(np.array(list1),axis = 1)
print(means)
结果：
[ 1.90367248 1.25196234 1.07593996 1.248519 1.08178562 1.25656678
1.08446815 1.26876521 1.09597256]
编辑:
如果你想对列进行平均，它是
means = np.mean(np.array(list1),axis = 0)
给出：
[ 0.68679585 0.34464285 0.20140261 5.83448231 0.44444444 0. ]
这应该有效
list1 = np.array(list1)
mean_col = list1[:,col_index].mean()
Column_index代表要计算均值的索引列，即column1的index = 0 ,column2=1.
我自己试过，有效:)
您在使用 numpy 时遇到的问题是示例中矩阵的声明。
给定：
list1 = [(4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)]
您可以轻松地使用它在 numpy 中按列获取平均值：
>>> np.mean(list1, axis=0)
[ 0.68679585 0.34464285 0.20140261 5.83448231 0.44444444 0. ]
你接下来有一个有趣的声明：
a = np.array([ (4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)],
dtype=[('col1', '<f8'), ('col2', '<f8'), ('col3', '<f8'), ('col4', '<f8'), ('col5', '<i4'), ('col6', '<i4')])
这与 matrix=np.array(list1) 不同它所做的是声明一个 numpy structured array 并命名每一列并为该列提供 dtype
该数组的每一行元素都是一个元组：
>>> a[0]
( 4.97487413, 0.43849328, 0.18793185, 5.82073561, 0, 0)
并且您不能以通常的方式访问列：
>>> a[:,0]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: too many indices for array
由于实际上是一维数组：
>>> a.shape
(9,)
相反，您必须按名称访问列：
>>> a['col1']
array([ 4.97487413, 0.15069597, 0.15085613, 0.15069597, 0.1509362 ,
0.15069597, 0.15085613, 0.1509362 , 0.1506159 ])
或者，按列名取平均值：
>>> [np.mean(a[col]) for col in ['col{}'.format(i) for i in range(1,7)]]
[0.68679584555500162, 0.34464284907500159, 0.20140260920884526, 5.8344823121106151, 0.44444444444444442, 0.0]
我有一个包含以下内容的列表：
list1 = [(4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)]
我需要这些值在垂直轴上的平均值，这样输出看起来像：
[(average_col1, average_col2, average_col3, average_col4, average_col5, average_col6)]
然而，np.mean(list1, axis=1)命令returns:
IndexError: tuple index out of range
因此我尝试使用以下方法创建一个 numpy 数组：
a = np.array(list1)
a = array([ (4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)],
dtype=[('col1', '<f8'), ('col2', '<f8'), ('col3', '<f8'), ('col4', '<f8'), ('col5', '<i4'), ('col6', '<i4')])
如果我使用与上面相同的平均命令 returns:
IndexError: tuple index out of range
因此我不确定从这里开始做什么。
您可以在不使用 numpy 的情况下为您的第一个列表尝试此操作：
averages = [sum(i)/float(len(i)) for i in zip(*list)]
您在尝试调用 np.mean 时使用了括号 ([]) 而不是圆括号 (())。这段代码应该做你想做的事：
import numpy as np
list1 = [(4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)]
means = np.mean(np.array(list1),axis = 1)
print(means)
结果：
[ 1.90367248 1.25196234 1.07593996 1.248519 1.08178562 1.25656678
1.08446815 1.26876521 1.09597256]
编辑:
如果你想对列进行平均，它是
means = np.mean(np.array(list1),axis = 0)
给出：
[ 0.68679585 0.34464285 0.20140261 5.83448231 0.44444444 0. ]
这应该有效
list1 = np.array(list1)
mean_col = list1[:,col_index].mean()
Column_index代表要计算均值的索引列，即column1的index = 0 ,column2=1.
我自己试过，有效:)
您在使用 numpy 时遇到的问题是示例中矩阵的声明。
给定：
list1 = [(4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)]
您可以轻松地使用它在 numpy 中按列获取平均值：
>>> np.mean(list1, axis=0)
[ 0.68679585 0.34464285 0.20140261 5.83448231 0.44444444 0. ]
你接下来有一个有趣的声明：
a = np.array([ (4.974874129422414, 0.4384932775564907, 0.1879318517703546, 5.820735609514166, 0, 0),
(0.15069597326856923, 0.2961961688603689, 0.21595885700786707, 5.848923022691187, 1, 0),
(0.15085612758502492, 0.28850876174946627, 0.18977362640233908, 5.826501216543082, 0, 0),
(0.15069597326856923, 0.2887489932217097, 0.2176404773200905, 5.834028536994648, 1, 0),
(0.15093620474325167, 0.3005203353595069, 0.18961347208652674, 5.849643723630468, 0, 0),
(0.15069597326856923, 0.3235825566813912, 0.21515808543054254, 5.849964035159586, 1, 0),
(0.15085612758502492, 0.3520099475391594, 0.18937324061280378, 5.814569613228549, 0, 0),
(0.15093620474325167, 0.3860427394179732, 0.2174803230046498, 5.858131979266134, 1, 0),
(0.1506158961103403, 0.42768286128894817, 0.18969354924443318, 5.807843071967709, 0, 0)],
dtype=[('col1', '<f8'), ('col2', '<f8'), ('col3', '<f8'), ('col4', '<f8'), ('col5', '<i4'), ('col6', '<i4')])
这与 matrix=np.array(list1) 不同它所做的是声明一个 numpy structured array 并命名每一列并为该列提供 dtype
该数组的每一行元素都是一个元组：
>>> a[0]
( 4.97487413, 0.43849328, 0.18793185, 5.82073561, 0, 0)
并且您不能以通常的方式访问列：
>>> a[:,0]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: too many indices for array
由于实际上是一维数组：
>>> a.shape
(9,)
相反，您必须按名称访问列：
>>> a['col1']
array([ 4.97487413, 0.15069597, 0.15085613, 0.15069597, 0.1509362 ,
0.15069597, 0.15085613, 0.1509362 , 0.1506159 ])
或者，按列名取平均值：
>>> [np.mean(a[col]) for col in ['col{}'.format(i) for i in range(1,7)]]
[0.68679584555500162, 0.34464284907500159, 0.20140260920884526, 5.8344823121106151, 0.44444444444444442, 0.0]