如何在熊猫的数据上绘制直方图?

时间:2021-11-19 22:55:40

I have a simple dataframe in pandas that has two numeric columns. I want to make a histogram out of the columns using matplotlib through pandas. The example below does not work:

我在熊猫中有一个简单的dataframe,它有两个数字列。我想用matplotlib在熊猫里做一个直方图。下面的例子不起作用:

In [6]: pandas.__version__
Out[6]: '0.14.1'

In [7]: df
Out[7]: 
   a   b
0  1  20
1  2  40
2  3  30
3  4  30
4  4   3
5  3   5

In [8]: df.plot(kind="hist")
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-8-4f53176a4683> in <module>()
----> 1 df.plot(kind="hist")

/software/lib/python2.7/site-packages/pandas/tools/plotting.pyc in plot_frame(frame, x, y, subplots, sharex, sharey, use_index, figsize, grid, legend, rot, ax, style, title, xlim, ylim, logx, logy, xticks, yticks, kind, sort_columns, fontsize, secondary_y, **kwds)
   2095         klass = _plot_klass[kind]
   2096     else:
-> 2097         raise ValueError('Invalid chart type given %s' % kind)
   2098 
   2099     if kind in _dataframe_kinds:

ValueError: Invalid chart type given hist

why does it say invalid chart type? the columns are numeric and can be made into histograms.

为什么说无效的图表类型?这些列是数字的,可以做成直方图。

2 个解决方案

#1


14  

DataFrame has its own hist method:

DataFrame有自己的hist方法:

df =pd.DataFrame({'col1':np.random.randn(100),'col2':np.random.randn(100)})
df.hist(layout=(1,2))   

draws a histogram for each valid column of the dataframe.

为dataframe的每个有效列绘制直方图。

如何在熊猫的数据上绘制直方图?

#2


2  

I don't believe 'hist' was a supported type in 0.14.1. Try df.hist() instead

我不认为hist是0.14.1中支持的类型。尝试df.hist()

#1


14  

DataFrame has its own hist method:

DataFrame有自己的hist方法:

df =pd.DataFrame({'col1':np.random.randn(100),'col2':np.random.randn(100)})
df.hist(layout=(1,2))   

draws a histogram for each valid column of the dataframe.

为dataframe的每个有效列绘制直方图。

如何在熊猫的数据上绘制直方图?

#2


2  

I don't believe 'hist' was a supported type in 0.14.1. Try df.hist() instead

我不认为hist是0.14.1中支持的类型。尝试df.hist()