How to replace dates in a column in a pandas dataframe?

Question

I want to fill a column with particular dates. First, I create a column of dates filled only with the date of today. Then, I compute the date I want to replace for each location. The index is in the following format: '%Y-%m-%d'

The computation works well. It is the last line which does not work

There is my code:

for j in range(1, 154):
    result['LTD_' + str(j)] = dt.date(dt.datetime.now().year,dt.datetime.now().month,dt.datetime.now().day) #creating a column of dates filled with today day
    for i in range(0, len(result.index)):
        date_selected = result.iloc[[i]].index
        date_month = add_months(date_selected,j)
        delivery_month = date_month.month
        delivery_year = date_month.year
        delivery = dt.date(delivery_year, delivery_month, result[(result.index.month == (delivery_month)) & (result.index.year == delivery_year)].index.day.min())
        delloc = result.index.get_loc(str(delivery))
        ltdloc = delloc - 3
        result.iloc[[i]]['LTD_' + str(j)] = dt.date(result.iloc[[ltdloc]].index.year, result.iloc[[ltdloc]].index.month, result.iloc[[ltdloc]].index.day)

The last line does not work to replace the today date by the computed date. Nothing happened. date.replace generates an error.

What is error? If use instaed last row result.iloc[[i]]['LTD_' + str(j)] = 0 it failed? If use print(dt.date(result.iloc[[ltdloc]].index.year, result.iloc[[ltdloc]].index.month, result.iloc[[ltdloc]].index.day)) it failed? — jezrael
– jezrael, Commented May 25, 2017 at 12:01
result.iloc[[i]]['LTD_' + str(j)] = 0, nothing happened. I have still the today date (2017-05-25). However, print(dt.date(result.iloc[[ltdloc]].index.year, result.iloc[[ltdloc]].index.month, result.iloc[[ltdloc]].index.day)) gives the good date in the format '%Y-%m-%d' I want to replace in result.iloc[[i]]['LTD_' + str(j)]. — Etienne
– Etienne, Commented May 25, 2017 at 12:36
Hmmm, without data it is really hard, but result['LTD_' + str(j)].iloc[i] or result['LTD_' + str(j)].iloc[[i]] should works. — jezrael
– jezrael, Commented May 25, 2017 at 12:39

jezrael · Accepted Answer · 2017-05-25 13:11:42Z

1

Now is ix deprecated, so need loc with selecting index by position:

result.loc[result.index[i], 'LTD_' + str(j)] = ...

Or iloc with get_loc for position of column name:

result.iloc[i, result.columns.get_loc('LTD_' + str(j))] = ...

Another solution is:

result['LTD_' + str(j)].iloc[[i]] = ...

edited May 25, 2017 at 13:11

answered May 25, 2017 at 12:43

jezrael

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2 Comments

Etienne Over a year ago

result['LTD_' + str(j)].iloc[[i]] solved my issue. Thank you very much jezrael.

jezrael Over a year ago

I add it to answer ;)

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