时间轮在python3熊猫

Question

如何使用登录/注销事件时间创建类似于下面的时间轮？特别希望以时间轮方式关联与星期几相关的平均登录/注销时间？下面的图片就是一个例子，但我正在寻找时间昼夜不停的时间，一周中的时间现在在图片中。我有可用的python和包含登录时间的数据集。我还想将颜色与用户类型相关联，例如管理员与普通用户或某种性质的用户。任何关于如何实现这一点的想法都会很棒。

一些示例数据位于pandas数据框中

DF：

TimeGenerated        EventID  Username  Message
2012-04-01 00:00:13  4624     Matthew   This guy logged onto the computer for the first time today
2012-04-01 00:00:14  4624     Matthew   This guy authenticated for some stuff 
2012-04-01 00:00:15  4624     Adam      This guy logged onto the computer for the first time today
2012-04-01 00:00:16  4624     James     This guy logged onto the computer for the first time today
2012-04-01 12:00:17  4624     Adam      This guy authenticated for some stuff
2012-04-01 12:00:18  4625     James     This guy logged off the computer for the last time today
2012-04-01 12:00:19  4624     Adam      This guy authenticated for some stuff
2012-04-01 12:00:20  4625     Adam      This guy logged off the computer for the last time today 
2012-04-01 12:00:21  4625     Matthew   This guy logged off the computer for the last time today

Answer 1

基本上，你需要完成2个不相交的任务：

• 创建一个可视化的频率表
• 定义一个可视化给定表的函数

对于第一项任务，我假设您只需要一个工作日和工作时间的数据透视表。我生成一个随机的：

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import matplotlib as mpl
import matplotlib.cm as cm
import calendar

# generate the table with timestamps
np.random.seed(1)
times = pd.Series(pd.to_datetime("Nov 1 '16 at 0:42") + pd.to_timedelta(np.random.rand(10000)*60*24*40, unit='m'))
# generate counts of each (weekday, hour)
data = pd.crosstab(times.dt.weekday, times.dt.hour.apply(lambda x: '{:02d}:00'.format(x))).fillna(0)
data.index = [calendar.day_name[i][0:3] for i in data.index]
print(data.T)

看起来像这样。每个号码都是此时登录的计数器：

       Mon  Tue  Wed  Thu  Fri  Sat  Sun
col_0                                   
00:00   55   56   67   60   60   62   45
01:00   51   65   70   65   60   59   40
02:00   47   76   67   68   61   63   51
....

现在，让我们为这张桌子绘制轮子！它将包含多个饼图：

# make a heatmap building function 
def pie_heatmap(table, cmap=cm.hot, vmin=None, vmax=None,inner_r=0.25, pie_args={}):
    n, m = table.shape
    vmin= table.min().min() if vmin is None else vmin
    vmax= table.max().max() if vmax is None else vmax

    centre_circle = plt.Circle((0,0),inner_r,edgecolor='black',facecolor='white',fill=True,linewidth=0.25)
    plt.gcf().gca().add_artist(centre_circle)
    norm = mpl.colors.Normalize(vmin=vmin, vmax=vmax)
    cmapper = cm.ScalarMappable(norm=norm, cmap=cmap)
    for i, (row_name, row) in enumerate(table.iterrows()):
        labels = None if i > 0 else table.columns
        wedges = plt.pie([1] * m,radius=inner_r+float(n-i)/n, colors=[cmapper.to_rgba(x) for x in row.values], 
            labels=labels, startangle=90, counterclock=False, wedgeprops={'linewidth':-1}, **pie_args)
        plt.setp(wedges[0], edgecolor='white',linewidth=1.5)
        wedges = plt.pie([1], radius=inner_r+float(n-i-1)/n, colors=['w'], labels=[row_name], startangle=-90, wedgeprops={'linewidth':0})
        plt.setp(wedges[0], edgecolor='white',linewidth=1.5)



plt.figure(figsize=(8,8))
pie_heatmap(data, vmin=-20,vmax=80,inner_r=0.2)

plt.show();

我希望这会对你有所帮助。

Answer 2

从@DavidDale的答案中获取数据，可以在极轴上绘制表格的pcolormesh图。这将直接给出所需的情节。

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
import calendar

# generate the table with timestamps
np.random.seed(1)
times = pd.Series(pd.to_datetime("Nov 1 '16 at 0:42") + 
                  pd.to_timedelta(np.random.rand(10000)*60*24*40, unit='m'))
# generate counts of each (weekday, hour)
data = pd.crosstab(times.dt.weekday, 
                   times.dt.hour.apply(lambda x: '{:02d}:00'.format(x))).fillna(0)
data.index = [calendar.day_name[i][0:3] for i in data.index]
data = data.T

# produce polar plot
fig, ax = plt.subplots(subplot_kw=dict(projection='polar'))
ax.set_theta_zero_location("N")
ax.set_theta_direction(-1)

# plot data
theta, r = np.meshgrid(np.linspace(0,2*np.pi,len(data)+1),np.arange(len(data.columns)+1))
ax.pcolormesh(theta,r,data.T.values, cmap="Reds")

# set ticklabels
pos,step = np.linspace(0,2*np.pi,len(data),endpoint=False, retstep=True)
pos += step/2.
ax.set_xticks(pos)
ax.set_xticklabels(data.index)

ax.set_yticks(np.arange(len(data.columns)))
ax.set_yticklabels(data.columns)
plt.show()