I have the column of DateTime which gets combined into one column. However, I would like to separate it into 2 columns of Date, Time.
The time is every fifteen minutes and I need to make it to every hour.
First off, make sure your datetime column is in datetime format
df['datetime'] = pd.to_datetime(df['datetime'])
You can then easily extract the date and hour from this using:
df['Date'] = df['datetime'].dt.date
df['Hour'] = df['datetime'].dt.hour
Related
I am trying to use this, but eventually, I get the same year-month-day format where my year changed to default "1900". I want to get only month-day pairs if it is possible.
df['date'] = pd.to_datetime(df['date'], format="%m-%d")
If you transform anything to date time, you'll always have a year in it, i.e. to_datetime will always yield a date time with a year.
Without a year, you will need to store it as a string, e.g. by running the inverse of your example:
df['date'] = df['date'].dt.strftime(format="%m-%d")
So I have sales data that I'm trying to analyze. I have datetime data ["Order Date Time"] and I'd like to see the most common hours for sales but more importantly I'd like to see what minutes have NO sales.
I have been spinning my wheels for a while and I can't get my brain around a solution. Any help is greatly appreciated.
I import the data:
df = pd.read_excel ('Audit Period.xlsx')
print (df)
I clean up the data:
# Remove all columns except `applieddate` and null rows
time_df = df[df["Order Date Time"].notnull()]
# Ensure the index is still sequential
time_df = time_df[["Order Date Time"]].reset_index(drop=True)
# Select the first 10 rows
time_df.head(10)
I convert to datetime and I look at the month totals:
# Convert applieddate to datetime
time_df = time_df.copy()
time_df["Order Date Time"] = time_df["Order Date Time"].apply(pd.to_datetime)
time_df = time_df.set_index(time_df["Order Date Time"])
# Group by month
grouped = time_df.resample("M").count()
time_df = pd.DataFrame({"count": grouped.values.flatten()}, index=grouped.index)
time_df.head(10)
I try to group by hour but that gives me totals per day/hour rather than totals per hour like every order ever at noon, etc:
# Group by hour
grouped = time_df.resample("2H").count()
time_df = pd.DataFrame({"count": grouped.values.flatten()}, index=grouped.index)
time_df.head(10)
And that is where I'm stuck. I'm trying to integrate the below suggestions but can't quite get a grasp on them yet. Any help would be appreciated.
Not sure if this is the most brilliant solution, but I would start by generating a dataframe at the level of detail I wanted, whether that is 1-hour intervals, 5-minute intervals, etc. Then in your df with all the actual data, you could do your grouping as you currently are doing it above. Once it is grouped, join the two. That way you have one dataframe that includes empty rows associated with time spans with no records. The tricky part will just be making sure you have your date and time formatted in a way that it will match and join properly.
I'm a beginner in python. I have an excel file. This file shows the rainfall amount between 2016-1-1 and 2020-6-30. It has 2 columns. The first column is date, another column is rainfall. Some dates are missed in the file (The rainfall didn't estimate). For example there isn't a row for 2016-05-05 in my file. This a sample of my excel file.
Date rainfall (mm)
1/1/2016 10
1/2/2016 5
.
.
.
12/30/2020 0
I want to find the missing dates but my code doesn't work correctly!
import pandas as pd
from datetime import datetime, timedelta
from matplotlib import dates as mpl_dates
from matplotlib.dates import date2num
df=pd.read_excel ('rainfall.xlsx')
a= pd.date_range(start = '2016-01-01', end = '2020-06-30' ).difference(df.index)
print(a)
Here' a beginner friendly way of doing it.
First you need to make sure, that the Date in your dataframe is really a date and not a string or object.
Type (or print) df.info().
The date column should show up as datetime64[ns]
If not, df['Date'] = pd.to_datetime(df['Date'], dayfirst=False)fixes that. (Use dayfirst to tell if the month is first or the day is first in your date string because Pandas doesn't know. Month first is the default, if you forget, so it would work without...)
For the tasks of finding missing days, there's many ways to solve it. Here's one.
Turn all dates into a series
all_dates = pd.Series(pd.date_range(start = '2016-01-01', end = '2020-06-30' ))
Then print all dates from that series which are not in your dataframe "Date" column. The ~ sign means "not".
print(all_dates[~all_dates.isin(df['Date'])])
Try:
df = pd.read_excel('rainfall.xlsx', usecols=[0])
a = pd.date_range(start = '2016-01-01', end = '2020-06-30').difference([l[0] for l in df.values])
print(a)
And the date in the file must like 2016/1/1
To find the missing dates from a list, you can apply Conditional Formatting function in Excel. 4. Click OK > OK, then the position of the missing dates are highlighted. Note: The last date in the date list will be highlighted.
this TRICK Is not with python,a NORMAL Trick
I have a pandas array with a column which contains unix timestamp times, but I think it's in milliseconds because each time as 3 extra 0's at the end. For example, the first data point is 1546300800000, when it should be just 1546300800. I need to convert this column to readable times so right now I have:
df = pd.read_csv('data.csv')
df['Time] = pd.to_datetime(df['Time'])
df.to_csv('data.csv', index=False)
Instead of giving me the correct time it gives me a time in 1970. For example 1546300800000 gives me 1970-01-01 00:25:46.301100 when it should be 2019-01-01 00:00:00. It does this for every timestamp in the column, which is over 20K rows
Data;
df=pd.DataFrame({'UNIX':['1349720105','1546300800']})
Conversion
df['UNIX']=pd.to_datetime(df['UNIX'], unit='s')
I have concatenated several csv files into one dataframe to make a combined csv file. But one of the columns has both date and time (e.g 02:33:01 21-Jun-2018) after being converted to date_time format. However when I call
new_dataframe = old_dataframe.sort_values(by = 'Time')
It sorts the dataframe by time , completely ignoring date.
Index Time Depth(ft) Pit Vol(bbl) Trip Tank(bbl)
189147 00:00:00 03-May-2018 2283.3578 719.6753 54.2079
3875 00:00:00 07-May-2018 5294.7308 1338.7178 29.5781
233308 00:00:00 20-May-2018 8073.7988 630.7964 41.3574
161789 00:00:01 05-May-2018 122.2710 353.6866 58.9652
97665 00:00:01 01-May-2018 16178.8666 769.1328 66.0688
How do I get it to sort by dates and then times , so that Aprils days come first, and come in chronological order?
In order to sort with your date first and then time, your Time column should be in the right way Date followed by Time. Currently, it's opposite.
You can do this:
df['Time'] = df['Time'].str.split(' ').str[::-1].apply(lambda x: ' '.join(x))
df['Time'] = pd.to_datetime(df['Time'])
Now sort your df by Time like this:
df.sort_values('Time')