Reputation: 77
I am new to Python and I am trying to create a database that is similar to this using pandas.
Below a simplified version of my df:
Timestamp A B C
0 2013-02-01 1 0 0
1 2013-02-02 2 10 18
2 2013-02-03 3 0 19
3 2013-02-04 4 12 20
4 2013-02-05 0 13 21
5 2013-02-06 6 14 22
6 2013-02-07 7 15 23
7 2013-02-08 0 0 0
First thing I did is create a new empty data frame to store the data in using this code:
# Create frequent pattern source database
df_frequent_pattern = pd.DataFrame(columns = ["Start Time", "End Time", "Active Appliances"])
# Create start_time and end_time series using pd.date_range
df_frequent_pattern["Start Time"] = pd.date_range("2013-02-1", "2013-02-08", freq = "D")
df_frequent_pattern["End Time"] = pd.date_range("2013-02-2", "2013-02-09", freq = "D")
Which gives the output below:
Start Time End Time Active Appliances
0 2013-02-01 2013-02-02 NaN
1 2013-02-02 2013-02-03 NaN
2 2013-02-03 2013-02-04 NaN
3 2013-02-04 2013-02-05 NaN
4 2013-02-05 2013-02-06 NaN
5 2013-02-06 2013-02-07 NaN
6 2013-02-07 2013-02-08 NaN
7 2013-02-08 2013-02-09 NaN
Based on this and this Stack overflow post I wrote the following code to assign the appliances to the right time resolution:
# Add the data to the correct 'active' period based on interval and merge the active appliances in the "active appliances column"
# Row counter for the loop
rows = 8
for row in range(rows):
# Check if appliance is active during time resoltuion
if df_frequent_pattern["Start Time"] <= df["Timestamp"] | df["Timestamp" <= df_frequent_pattern["End Time"]:
# Add all the appliance active during the time resolution to the column as a string value (e.g. "A, B, C")
df_frequent_pattern["Active Appliances"] = df["A", "B", "C"].apply(lambda row: '_'.join(row.values.astype(str)), axis = 1)
Unfortunately the code doesn't work and I get the following error
df_frequent_pattern["Active Appliances"] = df["A", "B", "C"].apply(lambda row: '_'.join(row.values.astype(str)), axis = 1)
^
SyntaxError: invalid syntax
Yet, the '=' seems to be correctly placed based on the second post. Any idea on how I can use my df to get the expected results as illustrated above?
It should looks like this:
Start Time End Time Active Appliances
0 2013-02-01 2013-02-02 "A"
1 2013-02-02 2013-02-03 "A,B,C"
2 2013-02-03 2013-02-04 "A,C"
3 2013-02-04 2013-02-05 "A,B,C"
4 2013-02-05 2013-02-06 "A,B,C"
5 2013-02-06 2013-02-07 "A,B,C"
6 2013-02-07 2013-02-08 "A,B,C"
7 2013-02-08 2013-02-09 ""
Upvotes: 3
Views: 252
Reputation: 23099
Let's do this in a few steps.
First, let's make sure your Timestamp
is a datetime.
df['Timestamp'] = pd.to_datetime(df['Timestamp'])
Then we can create a new dataframe based on a min and max values of your timestamp.
df1 = pd.DataFrame({'start_time' : pd.date_range(df['Timestamp'].min(), df['Timestamp'].max())})
df1['end_time'] = df1['start_time'] + pd.DateOffset(days=1)
start_time end_time
0 2013-02-01 2013-02-02
1 2013-02-02 2013-02-03
2 2013-02-03 2013-02-04
3 2013-02-04 2013-02-05
4 2013-02-05 2013-02-06
5 2013-02-06 2013-02-07
6 2013-02-07 2013-02-08
7 2013-02-08 2013-02-09
Now we need to create a dataframe to merge onto your start_time
column.
Let's filter out any values that are less than 0 and create a list of active appliances:
df = df.set_index('Timestamp')
# the remaining columns MUST be integers for this to work.
# or you'll need to subselect them.
df2 = df.mask(df.le(0)).stack().reset_index(1).groupby(level=0)\
.agg(active_appliances=('level_1',list)).reset_index(0)
# change .agg(active_appliances=('level_1',list) >
# to .agg(active_appliances=('level_1',','.join)
# if you prefer strings.
Timestamp active_appliances
0 2013-02-01 [A]
1 2013-02-02 [A, B, C]
2 2013-02-03 [A, C]
3 2013-02-04 [A, B, C]
4 2013-02-05 [B, C]
5 2013-02-06 [A, B, C]
6 2013-02-07 [A, B, C]
Then we can merge:
final = pd.merge(df1,df2,left_on='start_time',right_on='Timestamp',how='left').drop('Timestamp',1)
start_time end_time active_appliances
0 2013-02-01 2013-02-02 [A]
1 2013-02-02 2013-02-03 [A, B, C]
2 2013-02-03 2013-02-04 [A, C]
3 2013-02-04 2013-02-05 [A, B, C]
4 2013-02-05 2013-02-06 [B, C]
5 2013-02-06 2013-02-07 [A, B, C]
6 2013-02-07 2013-02-08 [A, B, C]
7 2013-02-08 2013-02-09 NaN
Upvotes: 2