Styjun

I have a dataframe (pulled from a dictionary) that contains the weekdays of the current week as column headers, filled with people names within the rows

I.e.

dict = '2019-03-25': ['Bob', '', 'Joe'], '2019-03-26': ['Abel', '', ''], '2019-03-27': ['', '', ''], '2019-03-28': ['', 'Bob', ''], '2019-03-29': ['Richard', 'Joe', 'Fred']

df = pd.DataFrame(data=dict)

currentdate = datetime.datetime.today().strftime('%Y-%m-%d')

I would like to highlight all of the cells in the entire column where column header = currentdate and then convert this to html

I looked elsewhere, and believe this is a unique scenario, as all I could find online was related to highlighting based on the values of the cells themselves, e.g. value is greater or less than or contains specific text, but nothing related to the value within the column header itself.

EDIT:
Basically I have a calendar in excel, which I pull into a dataframe then convert it to HTML and send it out as an email. Below is the result I've gotten to (minus the highlights). I would like to create an if statement for formatting the html, so that it highlights the column yellow when header == current date.

Note that it's not a great practice to keep data as the column name. In this case, that means the date record. It also makes the code much more complicated.

Try this code snippet -- I've added comments along the way.

import pandas as pd

# note that I've replaced '' with None
d = '2019-04-30': ['Bob', None, 'Joe'], '2019-03-26': ['Abel', None, None], '2019-03-27': [None, None, None], '2019-03-28': [None, 'Bob', None], '2019-03-29': ['Richard', 'Joe', 'Fred']

# load items of dictionary as dataframe, don't use dates as column names, assign column names
df = pd.DataFrame(data=d.items(), columns = ['date','name_list'])

# convert string to datetime
# it's better to compare datetimes directly, and forces the type to datetime, in case of dirty data
# also using pandas datetime instead of the standard python library
df.date = pd.to_datetime(df.date)

# this is your filter or 'where' clause, to keep only records that match today
df = df.loc[df['date'] == pd.to_datetime('today')]

# unpack list as individual rows, see https://stackoverflow.com/a/12681217/2327328
df = pd.concat([pd.Series(row['date'], row['name_list']) 
 for _, row in df.iterrows()]).reset_index()


# display as html
print (df.to_html())

Provides as an output:

<table border="1" class="dataframe">
 <thead>
 <tr style="text-align: right;">
 <th></th>
 <th>index</th>
 <th>0</th>
 </tr>
 </thead>
 <tbody>
 <tr>
 <th>0</th>
 <td>Bob</td>
 <td>2019-04-30</td>
 </tr>
 <tr>
 <th>1</th>
 <td>None</td>
 <td>2019-04-30</td>
 </tr>
 <tr>
 <th>2</th>
 <td>Joe</td>
 <td>2019-04-30</td>
 </tr>
 </tbody>
</table>

which looks displays this: