Python Pandas Series if else箱形图 [英] Python Pandas Series if else box plot

问题描述

我有很多词典格式的数据，并且我尝试使用pandas根据IF ELSE语句打印字符串.在我的示例中，病假在dict和Pandas的秘密数据中组成了一些数据:

I have alot of data in a dictionary format and I am attempting to use pandas print a string based on an IF ELSE statement. For my example ill make up some data in dict and covert to Pandas:

df = pd.DataFrame(dict(a=[1.5,2.8,9.3],b=[7.2,3.3,4.9],c=[13.1,4.9,15.9],d=[1.1,1.9,2.9]))

df

这将返回:

    a   b   c   d
0   1.5 7.2 13.1 1.1
1   2.8 3.3 4.9 1.9
2   9.3 4.9 15.9 2.9

我的IF ELSE声明:

My IF ELSE statement:

for col in df.columns:
    if (df[col] < 4).any():
        print('Zone %s does not make setpoint' % col)
    else:
        print('Zone %s is Normal' % col)

返回:

Zone a does not make setpoint
Zone b does not make setpoint
Zone c is Normal
Zone d does not make setpoint

但是现在我想添加一个额外的东西来创建一个箱形图，在该图上我不设定设定值，并且还要对数据框进行设定的平均值.我知道这是熊猫系列，但是可以使用pandas.Series.plot.box()吗?

But now I want to add in an extra to create a box plot where I am not making setpoint and also average the data frame where it is making setpoint. I know this is pandas series, but can pandas.Series.plot.box() be used?

这是我在df.apply(lamba x:)函数中使用的IF ELSE语句，我被困于试图让盒形图在熊猫系列中工作...任何建议，我们将不胜感激！

This is my IF ELSE statement that I am using in a function with df.apply(lamba x:) and I am stuck trying to get the box box plot to work in pandas series... Any advice is greatly appreciated!

import matplotlib.pyplot as plt

def _print(x):
    if (x < 4).any():
        print('Zone %s does not make setpoint' % x.name)
        df.boxplot()
        plt.show()
    else:
        print('Zone %s is Normal' % x.name)
        print('The average is %s' % x.mean())

致电df.apply(lambda x: _print(x))

module 'matplotlib' has no attribute 'show'

推荐答案

您可以像df['a'].plot.box()一样调用pandas.Series.plot.box()来获取列a的箱线图.

Sure you can call pandas.Series.plot.box() like df['a'].plot.box() to get the boxplot of your column a.

为了满足您的问题，我会这样做:

To fit with your question I would have done this:

def _print(x):
    if (x < 4).any():
        print('Zone %s does not make setpoint' % x.name)
        df[x.name].plot.box() #call x.name to retrieve the column name
        plt.show()
        print(df[x.name].describe())
    else:
        print('Zone %s is Normal' % x.name)
        print('The average is %s' % x.mean())
    print('---')

df.apply(lambda x: _print(x))

下面在zone B和zone C的输出摘要中进行了说明.

Illustrated below extract of the output for zone B and zone C.

请注意，您可以添加.describe()以获得箱线图和其他统计信息描述(请参见

Note that you can add .describe() to get the boxplot and other stats description (see documentation).

不过，根据此处提出的解决方案，我将以不同的方式处理该问题.

Nevertheless I would have approach the problem differently, according to the solution proposed here.

另一种解决方案

您可以过滤数据框以拆分为make setpoint与否:

You can filter your dataframe to split into make setpoint or not:

s = df.apply(lambda x: not (x < 4).any())

然后在未设置设定值的框上绘制框.
如果变化不太大且区域不多，则在图中绘制所有图形:

Then plot the boxes on the one that doesn't make the set point.
Plot all in a figure if the variation is not too large, and if there is not so many zones:

df[s[~s].index].boxplot()
plt.show()

或将它们分开:

for col in s[~s].index:
    df[col].plot.box()
    plt.show()

在两种情况下，都将统计信息保存在dataframe中:

In both case get the statistics in a dataframe:

statdf = df[s[~s].index].describe()
print(statdf)

              a         b         d
count  3.000000  3.000000  3.000000
mean   4.533333  5.133333  1.966667
std    4.178915  1.960442  0.901850
min    1.500000  3.300000  1.100000
25%    2.150000  4.100000  1.500000
50%    2.800000  4.900000  1.900000
75%    6.050000  6.050000  2.400000
max    9.300000  7.200000  2.900000

通过这种方式，您可以通过statdf.loc['mean']获取统计信息(例如"mean").

This way you can get the stat (say 'mean' for instance) with statdf.loc['mean'].

如果要打印确实达到设定点的平均值:

If you want to print the mean of the one that does make the set point:

print(df[s[s].index].mean())

c    11.3
Name: mean, dtype: float64

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