1

I have the data frame:

dataframe

I want to plot the date column vs the close column.

Here is the code to do that:

import pandas as pd
import matplotlib.pyplot as plt



df=pd.read_csv('C:/Users/admin/Desktop/ohlcv.csv')

df['date']=pd.to_datetime(df['date'])
df=df.set_index('date')


plt.figure(figsize=(12.5,4.5))
plt.title('Close Price Variation')
plt.xlabel('date')
plt.ylabel('close')
plt.plot(df.close)
plt.show()

This gives the plot as shown:

Plot

The issue is that the timestamp in the original plot is not the same as in the dataframe. What I mean by this is that data in the dataframe is spaced by 15 mins interval while in plot is spaced in 1 min interval

How can I get the same timestamp in the plot and in the dataframe

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7
  • 1
    Can you show the entire date column in your table? It is crucial to see the formatting Commented Mar 9, 2021 at 7:21
  • @Shir Added the required column Commented Mar 9, 2021 at 7:24
  • I think the problem is with the +05:30 suffix in every date. Does it have any significance to you? I'd try removing it and plotting without it. I'll make the code snippet. Commented Mar 9, 2021 at 7:26
  • You can I just need the time like 9,9:15 and so on Commented Mar 9, 2021 at 7:26
  • 1
    @Huzefa So you need to rotate them. Commented Mar 9, 2021 at 8:11

2 Answers 2

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3

If the time series of your data contains a time zone, you can use tz_localize(None) to remove the time zone and make only the original time series. Next, we use the MinitueLocator to specify a 15-minute interval for the time series. Then we rotate the strings to avoid overlapping

import yfinance as yf
import pandas as pd

df = yf.download("AAPL", interval='15m', start="2021-03-03", end="2021-03-04")

df.head(5)

    Open    High    Low     Close   Adj Close   Volume
Datetime                        
2021-03-02 10:00:00-05:00   126.735001  127.198402  126.541000  126.654999  126.654999  3163537
2021-03-02 10:15:00-05:00   126.650101  126.940002  125.980003  126.069901  126.069901  4519324
2021-03-02 10:30:00-05:00   126.059998  126.430000  125.779999  125.894997  125.894997  4357847
2021-03-02 10:45:00-05:00   125.879997  126.349998  125.730003  126.129997  126.129997  3172543
2021-03-02 11:00:00-05:00   126.129997  126.449997  125.849998  126.114998  126.114998  2971019

df.index = pd.to_datetime(df.index.tz_localize(None))

df.head(5)
    Open    High    Low     Close   Adj Close   Volume
Datetime                        
2021-03-02 10:00:00     126.735001  127.198402  126.541000  126.654999  126.654999  3163537
2021-03-02 10:15:00     126.650101  126.940002  125.980003  126.069901  126.069901  4519324
2021-03-02 10:30:00     126.059998  126.430000  125.779999  125.894997  125.894997  4357847
2021-03-02 10:45:00     125.879997  126.349998  125.730003  126.129997  126.129997  3172543
2021-03-02 11:00:00     126.129997  126.449997  125.849998  126.114998  126.114998  2971019

import matplotlib.pyplot as plt
import matplotlib.dates as mdates

fig = plt.figure(figsize=(12, 9))
ax = fig.add_subplot(111)
ax.set_title('Close Price Variation')
ax.set_xlabel('Date')
ax.set_ylabel('Close')
ax.plot(df.Close[:-2])

mins = mdates.MinuteLocator(byminute=[0,15,30,45])
ax.xaxis.set_major_locator(mins)

ax.tick_params(axis='x', labelrotation=45)
plt.show()

enter image description here

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Comments

2

You can correctly format the date column and this solves the problem:

import pandas as pd
import matplotlib.pyplot as plt

#df=pd.read_csv('C:/Users/admin/Desktop/ohlcv.csv')
df = pd.DataFrame({"date":["2021-03-09 09:15:00+05:30", 
                           "2021-03-09 09:30:00+05:30"],
                   "close":[615.8, 615.5]})
print(df.head())
df['date']=pd.to_datetime(df['date'], format="%Y-%m-%d %H:%M:%S%z")
df=df.set_index('date')

plt.figure(figsize=(12.5,4.5))
plt.title('Close Price Variation')
plt.xlabel('date')
plt.ylabel('close')
plt.plot(df.close)
plt.show()

Notice the %z in date formatting deals with the suffix discussed in the comments - UTC offset. For more information is strftime pandas documentation.

enter image description here

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2 Comments

Your answer is correct but does not work with large data. I have updated the dataframe to match the original length
@Huzefa I updated the solution such that the apply function redundant. This should work for data frames in any size. Simply load your own instead of the dummy frame I created.

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