python - Matplotlib(Seaborn) set_xticks 与日期时间和时间增量意外工作

Question

我应该先说这一切都是在 iPython 内核中完成的，但我采取的唯一操作是下面的代码。

我有以下由以下代码生成的图表:

from queries import TOTAL, DEMO, DB_CREDENTIALS, TOTAL_USA_EX, TOTAL_ESPN_EX
import pandas as pd
import pyodbc
pd.options.mode.chained_assignment = None  # default='warn'    
import seaborn as sns
from matplotlib import pyplot as plt
from datetime import datetime, timedelta
mpl.rc('font',family='Arial Rounded MT Bold')
y_label = {'fontsize':14}
title = {'fontsize':30}
s_legend = {'fontsize':14, 'handlelength':7}

with pyodbc.connect(DB_CREDENTIALS) as cnxn:
    df = pd.read_sql(sql=TOTAL_USA_EX, con=cnxn)
    df['date'] = pd.to_datetime(df['date'])
    df_e = pd.read_sql(sql=TOTAL_ESPN_EX, con=cnxn)
    df_e['date'] = pd.to_datetime(df_e['date'])

ex_ = df
ex_['subject'] = ex_['date'] - ex_['date'].min()
ex_['subject'] = ex_['subject'].apply(lambda x: x.days)
ex_['hour'] = ex_['datetime'].apply(lambda x: x.hour)
ex_['minute'] = ex_['datetime'].apply(lambda x: x.minute)
ex_['minute'] = ex_['minute'] // 15
ex_['qh'] = ex_.apply(lambda x: x['minute'] + (x['hour']*4), axis=1)
ex_['imp'] = ex_['imp'].apply(lambda x: round(x/1000000.0,3))
ex_['station'] = 'USA'

ex_e = df_e
ex_e['subject'] = ex_e['date'] - ex_e['date'].min()
ex_e['subject'] = ex_e['subject'].apply(lambda x: x.days)
ex_e['hour'] = ex_e['datetime'].apply(lambda x: x.hour)
ex_e['minute'] = ex_e['datetime'].apply(lambda x: x.minute)
ex_e['minute'] = ex_e['minute'] // 15
ex_e['qh'] = ex_e.apply(lambda x: x['minute'] + (x['hour']*4), axis=1)
ex_e['imp'] = ex_e['imp'].apply(lambda x: round(x/1000000.0,3))
ex_e['station'] = 'ESPN'

data = pd.concat([ex_, ex_e])        

fig, ax = plt.subplots()
fig.set_size_inches(14, 7)
sns.tsplot(time='qh', value='imp', unit='subject', condition='station', 
           ci=80, data=data, ax=ax, linewidth=2, color=["#21A0A0", "#E53D00"])
ax.set_ylabel('IMPRESSIONS (M)', **y_label)
ax.set_xlabel('TIME', **y_label)
ax.set_title('STATION IMPRESSIONS: 80% CONFIDENCE INTERVAL')
ax.set_xticks([x for x in xrange(0,96,8)])
ax.set_xticklabels([(datetime(year=2015,month=12,day=28)+timedelta(minutes=15*(x))).strftime('%H:%M') for x in ax.get_xticks()]);

x_ticks 设置为 15 分钟间隔，因此预期行为是每 2 小时增量设置一次刻度(例如 xticklabel[0] = 00:00、xticklabel[1 ] = 02:00，依此类推)。

但是，由于某种原因，会产生以下结果:

我在下面添加了日期和月份，以查看到底发生了什么，但仍然令人困惑。

因此，我直观地尝试通过查看创建 ax 后尝试访问刻度对象时发生的情况并查看计算是否完成来重新创建错误，它揭示了一些 super 令人困惑的内容行为:

In [19]: 
i = ax.get_xticks()
[(timedelta(minutes=15*(j)), j) for j in i ]

Out [19]:
[(datetime.timedelta(0), 0),
 (datetime.timedelta(-1, 85010, 65408), 8),
 (datetime.timedelta(0, 1515, 98112), 16),
 (datetime.timedelta(0, 125, 163520), 24),
 (datetime.timedelta(-1, 85135, 228928), 32),
 (datetime.timedelta(0, 1640, 261632), 40),
 (datetime.timedelta(0, 250, 327040), 48),
 (datetime.timedelta(-1, 85260, 392448), 56),
 (datetime.timedelta(0, 1765, 425152), 64),
 (datetime.timedelta(0, 375, 490560), 72),
 (datetime.timedelta(-1, 85385, 555968), 80),
 (datetime.timedelta(0, 1890, 588672), 88)]

为了我的理智，i 是什么？

In [20]:
i
Out [20]:
array([ 0,  8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88])

因此，我在 jupyter 中打开一个单独的内核，看看是否可以在真空中复制相同的错误。我无法:

新内核

In [1]:
from datetime import datetime, timedelta
i = [x*15 for x in xrange(0,96,8)]
[timedelta(minutes=x) for x in i]

Out [1]:
[datetime.timedelta(0),
 datetime.timedelta(0, 7200),
 datetime.timedelta(0, 14400),
 datetime.timedelta(0, 21600),
 datetime.timedelta(0, 28800),
 datetime.timedelta(0, 36000),
 datetime.timedelta(0, 43200),
 datetime.timedelta(0, 50400),
 datetime.timedelta(0, 57600),
 datetime.timedelta(0, 64800),
 datetime.timedelta(0, 72000),
 datetime.timedelta(0, 79200)]

有人可以帮助我不要发疯吗？

两次快速编辑:

1) 日期 12-28-2015 完全是任意的，我不需要该日期，我只需要与我的轴相关的时间。任何日期都可以，但考虑到我所期望的行为，这应该不重要。

2)只是为了确保它不是某种奇怪的语法错误，类似地在新内核中它工作得很好:

In [2]:
from datetime import datetime, timedelta
i = [x for x in xrange(0,96,8)]
[timedelta(minutes=(x)*15) for x in i]
Out [2]:
[datetime.timedelta(0),
 datetime.timedelta(0, 7200),
 datetime.timedelta(0, 14400),
 datetime.timedelta(0, 21600),
 datetime.timedelta(0, 28800),
 datetime.timedelta(0, 36000),
 datetime.timedelta(0, 43200),
 datetime.timedelta(0, 50400),
 datetime.timedelta(0, 57600),
 datetime.timedelta(0, 64800),
 datetime.timedelta(0, 72000),
 datetime.timedelta(0, 79200)]

Answer 1

感谢 #learnprogramming channel 中的 darkf；这是由 ax.get_xticks() 方法返回的项目类型引起的，即 numpy.int32；因此它可能返回一个指针引用而不是实际的 int。

更正的代码行:

x.set_xticklabels([(datetime(year=2015,month=12,day=28)+timedelta(minutes=15*(int(x)))).strftime('%H:%M') for x in ax.get_xticks()]);

和图表:

谢谢!