Python pandas-具有非数值的pivot_表？（DataError:没有要聚合的数字类型）

我正在尝试对包含字符串作为结果的表进行透视

import pandas as pd

df1 = pd.DataFrame({'index' : range(8),
'variable1' : ["A","A","B","B","A","B","B","A"],
'variable2' : ["a","b","a","b","a","b","a","b"],
'variable3' : ["x","x","x","y","y","y","x","y"],
'result': ["on","off","off","on","on","off","off","on"]})

df1.pivot_table(values='result',rows='index',cols=['variable1','variable2','variable3'])

但是我得到：

DataError:没有要聚合的数值类型

当我将结果值更改为数字时，这将按预期工作：

df2 = pd.DataFrame({'index' : range(8),
'variable1' : ["A","A","B","B","A","B","B","A"],
'variable2' : ["a","b","a","b","a","b","a","b"],
'variable3' : ["x","x","x","y","y","y","x","y"],
'result': [1,0,0,1,1,0,0,1]})

df2.pivot_table(values='result',rows='index',cols=['variable1','variable2','variable3'])

我得到了我需要的：

variable1   A               B    
variable2   a       b       a   b
variable3   x   y   x   y   x   y
index                            
0           1 NaN NaN NaN NaN NaN
1         NaN NaN   0 NaN NaN NaN
2         NaN NaN NaN NaN   0 NaN
3         NaN NaN NaN NaN NaN   1
4         NaN   1 NaN NaN NaN NaN
5         NaN NaN NaN NaN NaN   0
6         NaN NaN NaN NaN   0 NaN
7         NaN NaN NaN   1 NaN NaN

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我知道我可以将字符串映射为数值，然后反转操作，但也许有一个更优雅的解决方案？

我认为最好的折衷办法是用真/假替换开/关，这将使熊猫更好地“理解”数据，并以智能、预期的方式行事

df2 = df1.replace({'on': True, 'off': False})

你在问题中基本上承认了这一点。我的回答是，我不认为有更好的方法，不管接下来发生什么，你都应该替换“开”/“关”

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正如Andy Hayden在评论中指出的那样，如果用1/0替换开/关，您将获得更好的性能。

我最初的回答基于Pandas 0.14.1，从那以后，pivot_表函数中的许多内容都发生了变化（行-->索引，列-->列…）

此外，我发布的原始lambda技巧似乎不再适用于熊猫0.18。您必须提供一个缩减函数（即使是最小值、最大值或平均值）。但即使这样似乎也不合适——因为我们不是在减少数据集，而是在转换它。。。。所以我更努力地看着unstack

import pandas as pd

df1 = pd.DataFrame({'index' : range(8),
'variable1' : ["A","A","B","B","A","B","B","A"],
'variable2' : ["a","b","a","b","a","b","a","b"],
'variable3' : ["x","x","x","y","y","y","x","y"],
'result': ["on","off","off","on","on","off","off","on"]})

# these are the columns to end up in the multi-index columns.
unstack_cols = ['variable1', 'variable2', 'variable3']

首先，使用index+要堆叠的列对数据设置索引，然后使用level arg调用unstack

df1.set_index(['index'] + unstack_cols).unstack(level=unstack_cols)

生成的数据帧如下所示

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+1，但是使用1和0可能更好，因为DataFrame具有float而不是object dtype:）最后一个替换pivot（）的解决方案在0.17中进行了更改。1@RandallGoodwin，我意识到这个问题已经两年了，但我用lambda得到了错误“ValueError:Function not reduce”，你会知道为什么吗？另一个想法是：如果你可能会出现多个值，你可以通过让你的

aggfunc=lambda x:：.join（[str（y）代表y in x]）

@dllahr使用相同的想法来压缩字符串，但你也可以使用各种字符串访问器。e、 g.

aggfunc=lambda x:x.str.cat（）

FWIW，我在这里用了一个答案：@RustyShackleford请看我或dllahr的评论