how to get ticks every hour?












8















Consider this simple example



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.ticker import FuncFormatter
import matplotlib.dates as mdates

pd.__version__
Out[147]: u'0.22.0'

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.Series(np.random.randn(len(idx)), index = idx)
df.head()
Out[145]:
2017-01-01 05:03:00 0.4361
2017-01-01 05:04:00 0.9737
2017-01-01 05:05:00 0.8430
2017-01-01 05:06:00 0.4292
2017-01-01 05:07:00 0.5739
Freq: T, dtype: float64


I want to plot this, and have ticks every hour. I use:



fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1) #
h_fmt = mdates.DateFormatter('%H:%M:%S')

df.plot(ax = ax, color = 'black', linewidth = 0.4)

ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)


which gives



enter image description here



why dont the ticks appear every hour here? Thanks for your help!










share|improve this question

























  • You should play with xtick.

    – user32185
    Feb 14 '18 at 15:32











  • do you know how to do it? thanks!!

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:34






  • 1





    @ℕʘʘḆḽḘ: so your desired output is just to see spikes at every full hour?

    – Skandix
    Feb 14 '18 at 15:36






  • 1





    possibly relevant: stackoverflow.com/questions/45704366/…

    – tmdavison
    Feb 14 '18 at 15:42











  • @Skandix, yes, I want to see ticks not only at 8, 10, 12 but every hour at 6,7,8,9,10 etc

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:44
















8















Consider this simple example



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.ticker import FuncFormatter
import matplotlib.dates as mdates

pd.__version__
Out[147]: u'0.22.0'

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.Series(np.random.randn(len(idx)), index = idx)
df.head()
Out[145]:
2017-01-01 05:03:00 0.4361
2017-01-01 05:04:00 0.9737
2017-01-01 05:05:00 0.8430
2017-01-01 05:06:00 0.4292
2017-01-01 05:07:00 0.5739
Freq: T, dtype: float64


I want to plot this, and have ticks every hour. I use:



fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1) #
h_fmt = mdates.DateFormatter('%H:%M:%S')

df.plot(ax = ax, color = 'black', linewidth = 0.4)

ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)


which gives



enter image description here



why dont the ticks appear every hour here? Thanks for your help!










share|improve this question

























  • You should play with xtick.

    – user32185
    Feb 14 '18 at 15:32











  • do you know how to do it? thanks!!

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:34






  • 1





    @ℕʘʘḆḽḘ: so your desired output is just to see spikes at every full hour?

    – Skandix
    Feb 14 '18 at 15:36






  • 1





    possibly relevant: stackoverflow.com/questions/45704366/…

    – tmdavison
    Feb 14 '18 at 15:42











  • @Skandix, yes, I want to see ticks not only at 8, 10, 12 but every hour at 6,7,8,9,10 etc

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:44














8












8








8


1






Consider this simple example



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.ticker import FuncFormatter
import matplotlib.dates as mdates

pd.__version__
Out[147]: u'0.22.0'

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.Series(np.random.randn(len(idx)), index = idx)
df.head()
Out[145]:
2017-01-01 05:03:00 0.4361
2017-01-01 05:04:00 0.9737
2017-01-01 05:05:00 0.8430
2017-01-01 05:06:00 0.4292
2017-01-01 05:07:00 0.5739
Freq: T, dtype: float64


I want to plot this, and have ticks every hour. I use:



fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1) #
h_fmt = mdates.DateFormatter('%H:%M:%S')

df.plot(ax = ax, color = 'black', linewidth = 0.4)

ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)


which gives



enter image description here



why dont the ticks appear every hour here? Thanks for your help!










share|improve this question
















Consider this simple example



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.ticker import FuncFormatter
import matplotlib.dates as mdates

pd.__version__
Out[147]: u'0.22.0'

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.Series(np.random.randn(len(idx)), index = idx)
df.head()
Out[145]:
2017-01-01 05:03:00 0.4361
2017-01-01 05:04:00 0.9737
2017-01-01 05:05:00 0.8430
2017-01-01 05:06:00 0.4292
2017-01-01 05:07:00 0.5739
Freq: T, dtype: float64


I want to plot this, and have ticks every hour. I use:



fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1) #
h_fmt = mdates.DateFormatter('%H:%M:%S')

df.plot(ax = ax, color = 'black', linewidth = 0.4)

ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)


which gives



enter image description here



why dont the ticks appear every hour here? Thanks for your help!







python pandas matplotlib






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 14 '18 at 15:26







ℕʘʘḆḽḘ

















asked Feb 14 '18 at 15:10









ℕʘʘḆḽḘℕʘʘḆḽḘ

6,940945102




6,940945102













  • You should play with xtick.

    – user32185
    Feb 14 '18 at 15:32











  • do you know how to do it? thanks!!

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:34






  • 1





    @ℕʘʘḆḽḘ: so your desired output is just to see spikes at every full hour?

    – Skandix
    Feb 14 '18 at 15:36






  • 1





    possibly relevant: stackoverflow.com/questions/45704366/…

    – tmdavison
    Feb 14 '18 at 15:42











  • @Skandix, yes, I want to see ticks not only at 8, 10, 12 but every hour at 6,7,8,9,10 etc

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:44



















  • You should play with xtick.

    – user32185
    Feb 14 '18 at 15:32











  • do you know how to do it? thanks!!

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:34






  • 1





    @ℕʘʘḆḽḘ: so your desired output is just to see spikes at every full hour?

    – Skandix
    Feb 14 '18 at 15:36






  • 1





    possibly relevant: stackoverflow.com/questions/45704366/…

    – tmdavison
    Feb 14 '18 at 15:42











  • @Skandix, yes, I want to see ticks not only at 8, 10, 12 but every hour at 6,7,8,9,10 etc

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 15:44

















You should play with xtick.

– user32185
Feb 14 '18 at 15:32





You should play with xtick.

– user32185
Feb 14 '18 at 15:32













do you know how to do it? thanks!!

– ℕʘʘḆḽḘ
Feb 14 '18 at 15:34





do you know how to do it? thanks!!

– ℕʘʘḆḽḘ
Feb 14 '18 at 15:34




1




1





@ℕʘʘḆḽḘ: so your desired output is just to see spikes at every full hour?

– Skandix
Feb 14 '18 at 15:36





@ℕʘʘḆḽḘ: so your desired output is just to see spikes at every full hour?

– Skandix
Feb 14 '18 at 15:36




1




1





possibly relevant: stackoverflow.com/questions/45704366/…

– tmdavison
Feb 14 '18 at 15:42





possibly relevant: stackoverflow.com/questions/45704366/…

– tmdavison
Feb 14 '18 at 15:42













@Skandix, yes, I want to see ticks not only at 8, 10, 12 but every hour at 6,7,8,9,10 etc

– ℕʘʘḆḽḘ
Feb 14 '18 at 15:44





@Skandix, yes, I want to see ticks not only at 8, 10, 12 but every hour at 6,7,8,9,10 etc

– ℕʘʘḆḽḘ
Feb 14 '18 at 15:44












1 Answer
1






active

oldest

votes


















11














The problem is that while pandas in general directly wraps the matplotlib plotting methods, this is not the case for plots with dates. As soon as dates are involved, pandas uses a totally different numerical representation of dates and hence also uses its own locators for the ticks.



In case you want to use matplotlib.dates formatters or locators on plots created with pandas you may use the x_compat=True option in pandas plots.



df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)


This allows to use the matplotlib.dates formatters or locators as shown below.
Else you may replace df.plot(ax = ax, color = 'black', linewidth = 0.4) by



ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)


Complete example:



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')
df = pd.Series(np.random.randn(len(idx)), index = idx)

fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')

ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)
#or use
df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)
#Then tick and format with matplotlib:
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






If the motivation to use pandas here is (as stated in the comments below) to be able to use secondary_y, the equivalent for matplotlib plots would be a twin axes twinx.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.DataFrame(np.cumsum(np.random.randn(len(idx), 2),0),
index = idx, columns=list("AB"))

fig, ax = plt.subplots()
ax.plot(df.index, df["A"], color = 'black')
ax2 = ax.twinx()
ax2.plot(df.index, df["B"], color = 'indigo')

hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






share|improve this answer


























  • thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:22











  • I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:27






  • 2





    I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:28






  • 2





    The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:30






  • 1





    Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

    – ImportanceOfBeingErnest
    Feb 19 '18 at 2:19











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









11














The problem is that while pandas in general directly wraps the matplotlib plotting methods, this is not the case for plots with dates. As soon as dates are involved, pandas uses a totally different numerical representation of dates and hence also uses its own locators for the ticks.



In case you want to use matplotlib.dates formatters or locators on plots created with pandas you may use the x_compat=True option in pandas plots.



df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)


This allows to use the matplotlib.dates formatters or locators as shown below.
Else you may replace df.plot(ax = ax, color = 'black', linewidth = 0.4) by



ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)


Complete example:



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')
df = pd.Series(np.random.randn(len(idx)), index = idx)

fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')

ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)
#or use
df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)
#Then tick and format with matplotlib:
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






If the motivation to use pandas here is (as stated in the comments below) to be able to use secondary_y, the equivalent for matplotlib plots would be a twin axes twinx.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.DataFrame(np.cumsum(np.random.randn(len(idx), 2),0),
index = idx, columns=list("AB"))

fig, ax = plt.subplots()
ax.plot(df.index, df["A"], color = 'black')
ax2 = ax.twinx()
ax2.plot(df.index, df["B"], color = 'indigo')

hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






share|improve this answer


























  • thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:22











  • I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:27






  • 2





    I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:28






  • 2





    The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:30






  • 1





    Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

    – ImportanceOfBeingErnest
    Feb 19 '18 at 2:19
















11














The problem is that while pandas in general directly wraps the matplotlib plotting methods, this is not the case for plots with dates. As soon as dates are involved, pandas uses a totally different numerical representation of dates and hence also uses its own locators for the ticks.



In case you want to use matplotlib.dates formatters or locators on plots created with pandas you may use the x_compat=True option in pandas plots.



df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)


This allows to use the matplotlib.dates formatters or locators as shown below.
Else you may replace df.plot(ax = ax, color = 'black', linewidth = 0.4) by



ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)


Complete example:



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')
df = pd.Series(np.random.randn(len(idx)), index = idx)

fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')

ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)
#or use
df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)
#Then tick and format with matplotlib:
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






If the motivation to use pandas here is (as stated in the comments below) to be able to use secondary_y, the equivalent for matplotlib plots would be a twin axes twinx.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.DataFrame(np.cumsum(np.random.randn(len(idx), 2),0),
index = idx, columns=list("AB"))

fig, ax = plt.subplots()
ax.plot(df.index, df["A"], color = 'black')
ax2 = ax.twinx()
ax2.plot(df.index, df["B"], color = 'indigo')

hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






share|improve this answer


























  • thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:22











  • I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:27






  • 2





    I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:28






  • 2





    The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:30






  • 1





    Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

    – ImportanceOfBeingErnest
    Feb 19 '18 at 2:19














11












11








11







The problem is that while pandas in general directly wraps the matplotlib plotting methods, this is not the case for plots with dates. As soon as dates are involved, pandas uses a totally different numerical representation of dates and hence also uses its own locators for the ticks.



In case you want to use matplotlib.dates formatters or locators on plots created with pandas you may use the x_compat=True option in pandas plots.



df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)


This allows to use the matplotlib.dates formatters or locators as shown below.
Else you may replace df.plot(ax = ax, color = 'black', linewidth = 0.4) by



ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)


Complete example:



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')
df = pd.Series(np.random.randn(len(idx)), index = idx)

fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')

ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)
#or use
df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)
#Then tick and format with matplotlib:
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






If the motivation to use pandas here is (as stated in the comments below) to be able to use secondary_y, the equivalent for matplotlib plots would be a twin axes twinx.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.DataFrame(np.cumsum(np.random.randn(len(idx), 2),0),
index = idx, columns=list("AB"))

fig, ax = plt.subplots()
ax.plot(df.index, df["A"], color = 'black')
ax2 = ax.twinx()
ax2.plot(df.index, df["B"], color = 'indigo')

hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






share|improve this answer















The problem is that while pandas in general directly wraps the matplotlib plotting methods, this is not the case for plots with dates. As soon as dates are involved, pandas uses a totally different numerical representation of dates and hence also uses its own locators for the ticks.



In case you want to use matplotlib.dates formatters or locators on plots created with pandas you may use the x_compat=True option in pandas plots.



df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)


This allows to use the matplotlib.dates formatters or locators as shown below.
Else you may replace df.plot(ax = ax, color = 'black', linewidth = 0.4) by



ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)


Complete example:



import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')
df = pd.Series(np.random.randn(len(idx)), index = idx)

fig, ax = plt.subplots()
hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')

ax.plot(df.index, df.values, color = 'black', linewidth = 0.4)
#or use
df.plot(ax = ax, color = 'black', linewidth = 0.4, x_compat=True)
#Then tick and format with matplotlib:
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here






If the motivation to use pandas here is (as stated in the comments below) to be able to use secondary_y, the equivalent for matplotlib plots would be a twin axes twinx.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates

idx = pd.date_range('2017-01-01 05:03', '2017-01-01 18:03', freq = 'min')

df = pd.DataFrame(np.cumsum(np.random.randn(len(idx), 2),0),
index = idx, columns=list("AB"))

fig, ax = plt.subplots()
ax.plot(df.index, df["A"], color = 'black')
ax2 = ax.twinx()
ax2.plot(df.index, df["B"], color = 'indigo')

hours = mdates.HourLocator(interval = 1)
h_fmt = mdates.DateFormatter('%H:%M:%S')
ax.xaxis.set_major_locator(hours)
ax.xaxis.set_major_formatter(h_fmt)

fig.autofmt_xdate()
plt.show()


enter image description here







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share|improve this answer








edited Jun 6 '18 at 13:16

























answered Feb 14 '18 at 16:14









ImportanceOfBeingErnestImportanceOfBeingErnest

130k13139215




130k13139215













  • thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:22











  • I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:27






  • 2





    I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:28






  • 2





    The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:30






  • 1





    Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

    – ImportanceOfBeingErnest
    Feb 19 '18 at 2:19



















  • thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:22











  • I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

    – ℕʘʘḆḽḘ
    Feb 14 '18 at 16:27






  • 2





    I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:28






  • 2





    The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

    – ImportanceOfBeingErnest
    Feb 14 '18 at 16:30






  • 1





    Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

    – ImportanceOfBeingErnest
    Feb 19 '18 at 2:19

















thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

– ℕʘʘḆḽḘ
Feb 14 '18 at 16:22





thanks. maybe a workaround to keep Pandas doing the plots would be to get a regular Timestamp index instead of a pandas datetime?

– ℕʘʘḆḽḘ
Feb 14 '18 at 16:22













I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

– ℕʘʘḆḽḘ
Feb 14 '18 at 16:27





I would like to keep the Pandas constructor because I use extensively the secondary_y argument of plot to show multiple times series...

– ℕʘʘḆḽḘ
Feb 14 '18 at 16:27




2




2





I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

– ImportanceOfBeingErnest
Feb 14 '18 at 16:28





I can't imagine any workaround for this. matplotlib.dates assumes the numeric axes to represent days since 0001-01-01 UTC, plus 1. Only if the axes uses this datetime format, it will correctly tick and label the axes. On the other hand pandas will create its axes units depending on the data and use an appropriate locator for those units. The only alternative I could think of would be to manipulate the pandas formatters in use. I am not aware of any coherent way of doing this though.

– ImportanceOfBeingErnest
Feb 14 '18 at 16:28




2




2





The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

– ImportanceOfBeingErnest
Feb 14 '18 at 16:30





The equivalent of secondary_y in matplotlib is to create a twinx and plot the second plot to that axes. This will in general cause 3 codelines compared to one, so it's not that bad.

– ImportanceOfBeingErnest
Feb 14 '18 at 16:30




1




1





Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

– ImportanceOfBeingErnest
Feb 19 '18 at 2:19





Not sure what the problem is. Here the easiest would be fig.autofmt_xdate(rotation=90).

– ImportanceOfBeingErnest
Feb 19 '18 at 2:19


















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