Reputation: 4498
I have the following data frame
data_df =
date value
2016-01-15 1555
2016-01-16 1678
2016-01-17 1789
...
I would like to create a timeline graph, with the date as the x axis
I import the visualization modules
import matplotlib.pyplot as plt
%matplotlib inline
import vincent as vin
import seaborn as sb
I try to add a column to format the date data_df['dates'] = plt.date2num(ad_data.date)
Then I want to plot the timeline plot_date(data_df.dates, data_df.shown)
This doesn't work, since I am not converting the date correctly.
Upvotes: 2
Views: 11100
Reputation: 863541
You can use:
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.ticker as ticker
#sample data
start = pd.to_datetime('2016-01-15')
rng = pd.date_range(start, periods=100)
data_df = pd.DataFrame({'date': rng, 'value': range(100)})
data_df.value = data_df.value * 15 / data_df.date.dt.day
print (data_df)
date value
0 2016-01-15 0.000000
1 2016-01-16 0.937500
2 2016-01-17 1.764706
3 2016-01-18 2.500000
4 2016-01-19 3.157895
5 2016-01-20 3.750000
6 2016-01-21 4.285714
7 2016-01-22 4.772727
8 2016-01-23 5.217391
9 2016-01-24 5.625000
10 2016-01-25 6.000000
...
...
If necessary convert column date
to to_datetime
and then set_index
from column date
:
data_df.date = pd.to_datetime(data_df.date)
data_df.set_index('date', inplace=True)
print (data_df)
value
date
2016-01-15 0.000000
2016-01-16 0.937500
2016-01-17 1.764706
2016-01-18 2.500000
2016-01-19 3.157895
2016-01-20 3.750000
2016-01-21 4.285714
2016-01-22 4.772727
2016-01-23 5.217391
2016-01-24 5.625000
2016-01-25 6.000000
...
...
Plot Series
data_df['value']
by plot
and then set format of axis x
:
ax = data_df['value'].plot()
ticklabels = data_df.index.strftime('%Y-%m-%d')
ax.xaxis.set_major_formatter(ticker.FixedFormatter(ticklabels))
plt.show()
Upvotes: 5
Reputation: 6668
If your date is a datetime thingy (if not, use pd.to_datetime()
, it should recognize the format), it should work by just calling date_df.plot()
. Make sure it is set as index (so use date_df.index = date_df['date']
Upvotes: 0