绘制groupbys时，Seaborn出现“无法解释输入"错误 [英] 'Could not interpret input' error with Seaborn when plotting groupbys
问题描述
说我有这个数据框
d = { 'Path' : ['abc', 'abc', 'ghi','ghi', 'jkl','jkl'],
'Detail' : ['foo', 'bar', 'bar','foo','foo','foo'],
'Program': ['prog1','prog1','prog1','prog2','prog3','prog3'],
'Value' : [30, 20, 10, 40, 40, 50],
'Field' : [50, 70, 10, 20, 30, 30] }
df = DataFrame(d)
df.set_index(['Path', 'Detail'], inplace=True)
df
Field Program Value
Path Detail
abc foo 50 prog1 30
bar 70 prog1 20
ghi bar 10 prog1 10
foo 20 prog2 40
jkl foo 30 prog3 40
foo 30 prog3 50
我可以毫无问题地进行汇总(顺便说一句，如果有更好的方法可以做到这一点，我想知道！)
I can aggregate it no problem (if there's a better way to do this, by the way, I'd like to know!)
df_count = df.groupby('Program').count().sort(['Value'], ascending=False)[['Value']]
df_count
Program Value
prog1 3
prog3 2
prog2 1
df_mean = df.groupby('Program').mean().sort(['Value'], ascending=False)[['Value']]
df_mean
Program Value
prog3 45
prog2 40
prog1 20
我可以从Pandas绘制它，没问题...
I can plot it from Pandas no problem...
df_mean.plot(kind='bar')
但是为什么在seaborn中尝试时会出现此错误?
But why do I get this error when I try it in seaborn?
sns.factorplot('Program',data=df_mean)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-26-23c2921627ec> in <module>()
----> 1 sns.factorplot('Program',data=df_mean)
C:\Anaconda3\lib\site-packages\seaborn\categorical.py in factorplot(x, y, hue, data, row, col, col_wrap, estimator, ci, n_boot, units, order, hue_order, row_order, col_order, kind, size, aspect, orient, color, palette, legend, legend_out, sharex, sharey, margin_titles, facet_kws, **kwargs)
2673 # facets to ensure representation of all data in the final plot
2674 p = _CategoricalPlotter()
-> 2675 p.establish_variables(x_, y_, hue, data, orient, order, hue_order)
2676 order = p.group_names
2677 hue_order = p.hue_names
C:\Anaconda3\lib\site-packages\seaborn\categorical.py in establish_variables(self, x, y, hue, data, orient, order, hue_order, units)
143 if isinstance(input, string_types):
144 err = "Could not interperet input '{}'".format(input)
--> 145 raise ValueError(err)
146
147 # Figure out the plotting orientation
ValueError: Could not interperet input 'Program'
推荐答案
出现此异常的原因是 Program 成为数据帧 df_mean 的索引，并且 group_by 操作后的 df_count .
The reason for the exception you are getting is that Program becomes an index of the dataframes df_mean and df_count after your group_by operation.
如果您想从 df_mean 中获取 factorplot ，一个简单的解决方案是将索引添加为列，
If you wanted to get the factorplot from df_mean, an easy solution is to add the index as a column,
In [7]:
df_mean['Program'] = df_mean.index
In [8]:
%matplotlib inline
import seaborn as sns
sns.factorplot(x='Program', y='Value', data=df_mean)
不过，您甚至可以更简单地让 factorplot 为您进行计算，
However you could even more simply let factorplot do the calculations for you,
sns.factorplot(x='Program', y='Value', data=df)
您将获得相同的结果.希望对您有所帮助.
You'll obtain the same result. Hope it helps.
评论后编辑
实际上，您对参数 as_index 提出了非常好的建议；默认情况下，它设置为True，在这种情况下，就像您的问题一样， Program 成为索引的一部分.
Indeed you make a very good point about the parameter as_index; by default it is set to True, and in that case Program becomes part of the index, as in your question.
In [14]:
df_mean = df.groupby('Program', as_index=True).mean().sort(['Value'], ascending=False)[['Value']]
df_mean
Out[14]:
Value
Program
prog3 45
prog2 40
prog1 20
请清楚一点，这种方式 Program 不再是列，而是成为索引.技巧 df_mean ['Program'] = df_mean.index 实际上是保持索引不变，并为索引添加新列，以便现在复制 Program .
Just to be clear, this way Program is not column anymore, but it becomes the index. the trick df_mean['Program'] = df_mean.index actually keeps the index as it is, and adds a new column for the index, so that Program is duplicated now.
In [15]:
df_mean['Program'] = df_mean.index
df_mean
Out[15]:
Value Program
Program
prog3 45 prog3
prog2 40 prog2
prog1 20 prog1
但是，如果将 as_index 设置为False，则会将 Program 作为列，加上新的自动增量索引，
However, if you set as_index to False, you get Program as a column, plus a new autoincrement index,
In [16]:
df_mean = df.groupby('Program', as_index=False).mean().sort(['Value'], ascending=False)[['Program', 'Value']]
df_mean
Out[16]:
Program Value
2 prog3 45
1 prog2 40
0 prog1 20
这样，您可以将其直接喂入 seaborn .不过，您可以使用 df 并获得相同的结果.
This way you could feed it directly to seaborn. Still, you could use df and get the same result.
希望有帮助.
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