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Cannot select rows from pandas dataframe after converting date to month using function .dt.to_period('M')

I have a dataframe table like this:
df = pd.DataFrame({"txn_id":{'A','B','C'},"txn_date":{'2019-04-01','2020-06-01','2021-05-01'})
I was trying to find rows where transaction month is 2021-05
so what i did was:
import datetime
df['txn_month'] = df['txn_date'].dt.to_period('M')
df[df['txn_month'] == '2021-05']
However, the result returns nothing, even though in table i could see column "txn_month" has "2021-05"
could you please help? Thanks!

Use dt.to_period:
#convert to datetime if needed
df["txn_date"] = pd.to_datetime(df["txn_date"])
>>> df[df["txn_date"].dt.to_period("m").eq("2021-05")]
txn_id txn_date
2 C 2021-05-01

How to find missing dates in an excel file by python

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

Am i doing something wrong with the loops?

I am using python to do some data cleaning and i've used the datetime module to split date time and tried to create another column with just the time.
My script works but it just takes the last value of the data frame.
Here is the code:
import datetime
i = 0
for index, row in df.iterrows():
date = datetime.datetime.strptime(df.iloc[i, 0], "%Y-%m-%dT%H:%M:%SZ")
df['minutes'] = date.minute
i = i + 1
This is the dataframe :
Output

df['minutes'] = date.minute reassigns the entire 'minutes' column with the scalar value date.minute from the last iteration.
You don't need a loop, as 99% of the cases when using pandas.
You can use vectorized assignment, just replace 'source_column_name' with the name of the column with the source data.
df['minutes'] = pd.to_datetime(df['source_column_name'], format='%Y-%m-%dT%H:%M:%SZ').dt.minute
It is also most likely that you won't need to specify format as pd.to_datetime is fairly smart.
Quick example:
df = pd.DataFrame({'a': ['2020.1.13', '2019.1.13']})
df['year'] = pd.to_datetime(df['a']).dt.year
print(df)
outputs
a year
0 2020.1.13 2020
1 2019.1.13 2019

Seems like you're trying to get the time column from the datetime which is in string format. That's what I understood from your post.
Could you give this a shot?
from datetime import datetime
import pandas as pd
def get_time(date_cell):
dt = datetime.strptime(date_cell, "%Y-%m-%dT%H:%M:%SZ")
return datetime.strftime(dt, "%H:%M:%SZ")
df['time'] = df['date_time'].apply(get_time)

Convert a date to a different format for an entire new column

I want to convert the date in a column in a dataframe to a different format. Currently, it has this format: '2019-11-20T01:04:18'. I want it to have this format: 20-11-19 1:04.
I think I need to develop a loop and generate a new column for the new date format. So essentially, in the loop, I would refer to the initial column and then generate the variable for the new column in the format I want.
Can someone help me out to complete this task?
The following code works for one occasion:
import datetime
d = datetime.datetime.strptime('2019-11-20T01:04:18', '%Y-%m-%dT%H:%M:%S')
print d.strftime('%d-%m-%y %H:%M')

From a previous answer in this site , this should be able to help you, comments give explanation
You can read your data into pandas from csv or database or create some test data as shown below for testing.
>>> import pandas as pd
>>> df = pd.DataFrame({'column': {0: '26/1/2016', 1: '26/1/2016'}})
>>> # First convert the column to datetime datatype
>>> df['column'] = pd.to_datetime(df.column)
>>> # Then call the datetime object format() method, set the modifiers you want here
>>> df['column'] = df['column'].dt.strftime('%Y-%m-%dT%H:%M:%S')
>>> df
column
0 2016-01-26T00:00:00
1 2016-01-26T00:00:00
NB. Check to ensure that all your columns have similar date strings

You can either achieve it like this:
from datetime import datetime
df['your_column'] = df['your_column'].apply(lambda x: datetime.strptime(x, '%Y-%m-%dT%H:%M:%S').strftime('%d-%m-%y %H:%M'))

Pandas: select all dates with specific month and day

I have a dataframe full of dates and I would like to select all dates where the month==12 and the day==25 and add replace the zero in the xmas column with a 1.
Anyway to do this? the second line of my code errors out.
df = DataFrame({'date':[datetime(2013,1,1).date() + timedelta(days=i) for i in range(0,365*2)], 'xmas':np.zeros(365*2)})
df[df['date'].month==12 and df['date'].day==25] = 1

Pandas Series with datetime now behaves differently. See .dt accessor.
This is how it should be done now:
df.loc[(df['date'].dt.day==25) & (cust_df['date'].dt.month==12), 'xmas'] = 1

Basically what you tried won't work as you need to use the & to compare arrays, additionally you need to use parentheses due to operator precedence. On top of this you should use loc to perform the indexing:
df.loc[(df['date'].month==12) & (df['date'].day==25), 'xmas'] = 1

An update was needed in reply to this question. As of today, there's a slight difference in how you extract months from datetime objects in a pd.Series.
So from the very start, incase you have a raw date column, first convert it to datetime objects by using a simple function:
import datetime as dt
def read_as_datetime(str_date):
# replace %Y-%m-%d with your own date format
return dt.datetime.strptime(str_date,'%Y-%m-%d')
then apply this function to your dates column and save results in a new column namely datetime:
df['datetime'] = df.dates.apply(read_as_datetime)
finally in order to extract dates by day and month, use the same piece of code that #Shayan RC explained, with this slight change; notice the dt.datetime after calling the datetime column:
df.loc[(df['datetime'].dt.datetime.month==12) &(df['datetime'].dt.datetime.day==25),'xmas'] =1