Filling empty dataframe with loop

Question

I have the following dataframe

print(df.head(5))

                          date  places_occupees
0  2017-01-01 00:00:00.0000000              238
1  2017-01-01 00:01:00.0000000              238
2  2017-01-01 00:02:00.0000000              238
3  2017-01-01 00:03:00.0000000              238
4  2017-01-01 00:04:00.0000000              238
5  2017-01-01 00:05:00.0000000              238

(please note that the date column type is string)

I have a list of strings that I will use to sort data in the dataframe.

print(list_holidays)

['2017-01-01', '2017-05-01', '2017-05-08', '2017-07-14', '2017-11-11', '2017-04-17', '2017-06-05', '2017-05-25', '2017-08-15', '2017-11-01', '2017-12-25']

then I create a new empty dataframe with the same 2 columns. I will fill it with data using the loop:

new_df = pd.DataFrame(columns=['date', 'places_occupees'])

Here is what I used but returns an empty dataframe

for i in list_holidays:
    filter = df[df['date'].str.contains(i)]
    new_df['date'].append(filter.date)
    new_df['places_occupes'].append(filer.places_occupees)

What I would like to do is to fill the new_df 'date' column with the dates obtained after sorting and to fill the new_df 'places_occupees' column with the values obtained after sorting that should look like the initial dataframe but after applying a filter.

wait, why the need to create a dataframe and loop? what happens, for example, when there are multiple matches (like 2017-07-01) — zero
– zero, Commented Jul 16, 2019 at 10:10

Adam.Er8 · Accepted Answer · 2019-07-16 10:11:27Z

1

you can save your filter results (which are correct) in a list, then use pd.concat to get the new df.

try this:

filtered = []
for i in list_holidays:
    filter = df[df['date'].str.contains(i)]
    filtered.append(filter)

new_df = pd.concat(filtered)

print(new_df)

or with a simple list-comprehension:

new_df = pd.concat([df[df['date'].str.contains(i)] for i in list_holidays])

print(new_df)

answered Jul 16, 2019 at 10:11

Adam.Er8

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