Develop Reference

Combining two dataframes

I've tried merging two dataframes, but I can't seem to get it to work. Each time I merge, the rows where I expect values are all 0. Dataframe df1 already as some data in it, with some left blank. Dataframe df2 will populate those blank rows in df1 where column names match at each value in "TempBin" and each value in "Month" in df1.
EDIT:
Both dataframes are in a for loop. df1 acts as my "storage", df2 changes for each location iteration. So if df2 contained the results for LocationZP, I would also want that data inserted in the matching df1 rows. If I use df1 = df1.append(df2) in the for loop, all of the rows from df2 keep inserting at the very end of df1 for each iteration.
df1:
Month TempBin LocationAA LocationXA LocationZP
1 0 7 1 2
1 1 98 0 89
1 2 12 23 38
1 3 3 14 17
1 4 7 9 14
1 5 1 8 99
13 0 0 0 0
13 1 0 0 0
13 2 0 0 0
13 3 0 0 0
13 4 0 0 0
13 5 0 0 0
df2:
Month TempBin LocationAA
13 0 11
13 1 22
13 2 33
13 3 44
13 4 55
13 5 66
desired output in df1:
Month TempBin LocationAA LocationXA LocationZP
1 0 7 1 2
1 1 98 0 89
1 2 12 23 38
1 3 3 14 17
1 4 7 9 14
1 5 1 8 99
13 0 11 0 0
13 1 22 0 0
13 2 33 0 0
13 3 44 0 0
13 4 55 0 0
13 5 66 0 0
import pandas as pd
df1 = pd.DataFrame({'Month': [1]*6 + [13]*6,
'TempBin': [0,1,2,3,4,5]*2,
'LocationAA': [7,98,12,3,7,1,0,0,0,0,0,0],
'LocationXA': [1,0,23,14,9,8,0,0,0,0,0,0],
'LocationZP': [2,89,38,17,14,99,0,0,0,0,0,0]}
)
df2 = pd.DataFrame({'Month': [13]*6,
'TempBin': [0,1,2,3,4,5],
'LocationAA': [11,22,33,44,55,66]}
)
df1 = pd.merge(df1, df2, on=["Month","TempBin","LocationAA"], how="left")
result:
Month TempBin LocationAA LocationXA LocationZP
1 0 7.0 1.0 2.0
1 1 98.0 0.0 89.0
1 2 12.0 23.0 38.0
1 3 3.0 14.0 17.0
1 4 7.0 9.0 14.0
1 5 1.0 8.0 99.0
13 0 NaN NaN NaN
13 1 NaN NaN NaN
13 2 NaN NaN NaN
13 3 NaN NaN NaN
13 4 NaN NaN NaN
13 5 NaN NaN NaN

Here's some code that worked for me:
# Merge two df into one dataframe on the columns "TempBin" and "Month" filling nan values with 0.
import pandas as pd
df1 = pd.DataFrame({'Month': [1]*6 + [13]*6,
'TempBin': [0,1,2,3,4,5]*2,
'LocationAA': [7,98,12,3,7,1,0,0,0,0,0,0],
'LocationXA': [1,0,23,14,9,8,0,0,0,0,0,0],
'LocationZP': [2,89,38,17,14,99,0,0,0,0,0,0]}
)
df2 = pd.DataFrame({'Month': [13]*6,
'TempBin': [0,1,2,3,4,5],
'LocationAA': [11,22,33,44,55,66]})
df_merge = pd.merge(df1, df2, how='left',
left_on=['TempBin', 'Month'],
right_on=['TempBin', 'Month'])
df_merge.fillna(0, inplace=True)
# add column LocationAA and fill it with the not null value from column LocationAA_x and LocationAA_y
df_merge['LocationAA'] = df_merge.apply(lambda x: x['LocationAA_x'] if pd.isnull(x['LocationAA_y']) else x['LocationAA_y'], axis=1)
# remove column LocationAA_x and LocationAA_y
df_merge.drop(['LocationAA_x', 'LocationAA_y'], axis=1, inplace=True)
print(df_merge)
Output:
Month TempBin LocationXA LocationZP LocationAA
0 1 0 1.0 2.0 0.0
1 1 1 0.0 89.0 0.0
2 1 2 23.0 38.0 0.0
3 1 3 14.0 17.0 0.0
4 1 4 9.0 14.0 0.0
5 1 5 8.0 99.0 0.0
6 13 0 0.0 0.0 11.0
7 13 1 0.0 0.0 22.0
8 13 2 0.0 0.0 33.0
9 13 3 0.0 0.0 44.0
10 13 4 0.0 0.0 55.0
11 13 5 0.0 0.0 66.0
Let me know if there's something you don't understand in the comments :)
PS: Sorry for the extra comments. But I left them there for some more explanations.

You need to use append to get the desired output:
df1 = df1.append(df2)
and if you want to replace the Nulls to zeros add:
df1 = df1.fillna(0)

Here is another way using combine_first()
i = ['Month','TempBin']
df2.set_index(i).combine_first(df1.set_index(i)).reset_index()

combining two dataframes giving NaN values [duplicate]

I am curious why a simple concatenation of two dataframes in pandas:
initId.shape # (66441, 1)
initId.isnull().sum() # 0
ypred.shape # (66441, 1)
ypred.isnull().sum() # 0
of the same shape and both without NaN values
foo = pd.concat([initId, ypred], join='outer', axis=1)
foo.shape # (83384, 2)
foo.isnull().sum() # 16943
can result in a lot of NaN values if joined.
How can I fix this problem and prevent NaN values being introduced?
Trying to reproduce it like
aaa = pd.DataFrame([0,1,0,1,0,0], columns=['prediction'])
bbb = pd.DataFrame([0,0,1,0,1,1], columns=['groundTruth'])
pd.concat([aaa, bbb], axis=1)
failed e.g. worked just fine as no NaN values were introduced.

I think there is problem with different index values, so where concat cannot align get NaN:
aaa = pd.DataFrame([0,1,0,1,0,0], columns=['prediction'], index=[4,5,8,7,10,12])
print(aaa)
prediction
4 0
5 1
8 0
7 1
10 0
12 0
bbb = pd.DataFrame([0,0,1,0,1,1], columns=['groundTruth'])
print(bbb)
groundTruth
0 0
1 0
2 1
3 0
4 1
5 1
print (pd.concat([aaa, bbb], axis=1))
prediction groundTruth
0 NaN 0.0
1 NaN 0.0
2 NaN 1.0
3 NaN 0.0
4 0.0 1.0
5 1.0 1.0
7 1.0 NaN
8 0.0 NaN
10 0.0 NaN
12 0.0 NaN
Solution is reset_index if indexes values are not necessary:
aaa.reset_index(drop=True, inplace=True)
bbb.reset_index(drop=True, inplace=True)
print(aaa)
prediction
0 0
1 1
2 0
3 1
4 0
5 0
print(bbb)
groundTruth
0 0
1 0
2 1
3 0
4 1
5 1
print (pd.concat([aaa, bbb], axis=1))
prediction groundTruth
0 0 0
1 1 0
2 0 1
3 1 0
4 0 1
5 0 1
EDIT: If need same index like aaa and length of DataFrames is same use:
bbb.index = aaa.index
print (pd.concat([aaa, bbb], axis=1))
prediction groundTruth
4 0 0
5 1 0
8 0 1
7 1 0
10 0 1
12 0 1

You can do something like this:
concatenated_dataframes = concat(
[
dataframe_1.reset_index(drop=True),
dataframe_2.reset_index(drop=True),
dataframe_3.reset_index(drop=True)
],
axis=1,
ignore_index=True,
)
concatenated_dataframes_columns = [
list(dataframe_1.columns),
list(dataframe_2.columns),
list(dataframe_3.columns)
]
flatten = lambda nested_lists: [item for sublist in nested_lists for item in sublist]
concatenated_dataframes.columns = flatten(concatenated_dataframes_columns)
To concatenate multiple DataFrames and keep the columns names / avoid NaN.

As jezrael pointed out, this is due to different index labels. concat matches on index, so if they are not the same, this problem will occur. For a straightforward horizontal concatenation, you must "coerce" the index labels to be the same. One way is via set_axis method. This makes the second dataframes index to be the same as the first's.
joined_df = pd.concat([df1, df2.set_axis(df1.index)], axis=1)
or just reset the index of both frames
joined_df = pd.concat([df1.reset_index(drop=True), df2.reset_index(drop=True)], axis=1)

striding through pandas dataframe

I have a Dataframe of the form
date_time uids
2018-10-16 23:00:00 1000,1321,7654,1321
2018-10-16 23:10:00 7654
2018-10-16 23:20:00 NaN
2018-10-16 23:30:00 7654,1000,7654,1321,1000
2018-10-16 23:40:00 691,3974,3974,323
2018-10-16 23:50:00 NaN
2018-10-17 00:00:00 NaN
2018-10-17 00:10:00 NaN
2018-10-17 00:20:00 27,33,3974,3974,7665,27
This is a very big data frame containing the 5 mins time interval and the number of appearances of ids during those time intervals.
I want to iterate over these DataFrame 6 rows at a time (corresponding to 1 hour) and create DataFrame containing the ID and the number of times each id appear during this time.
Expected output is one dataframe per hour information. For example, in the above case dataframe for the hour 23 - 00 will have this form
uid 1 2 3 4 5 6
1000 1 0 0 2 0 0
1321 2 0 0 1 0 0
and so on
How can I do this efficiently?

I don't have an exact solution but you could create a pivot table: ids on the index and datetimes on the columns. Then you just have to select the columns you want.
import pandas as pd
import numpy as np
df = pd.DataFrame(
{
"date_time": [
"2018-10-16 23:00:00",
"2018-10-16 23:10:00",
"2018-10-16 23:20:00",
"2018-10-16 23:30:00",
"2018-10-16 23:40:00",
"2018-10-16 23:50:00",
"2018-10-17 00:00:00",
"2018-10-17 00:10:00",
"2018-10-17 00:20:00",
],
"uids": [
"1000,1321,7654,1321",
"7654",
np.nan,
"7654,1000,7654,1321,1000",
"691,3974,3974,323",
np.nan,
np.nan,
np.nan,
"27,33,3974,3974,7665,27",
],
}
)
df["date_time"] = pd.to_datetime(df["date_time"])
df = (
df.set_index("date_time") #do not use set_index if date_time is current index
.loc[:, "uids"]
.str.extractall(r"(?P<uids>\d+)")
.droplevel(level=1)
) # separate all the ids
df["number"] = df.index.minute.astype(float) / 10 + 1 # get the number 1 to 6 depending on the minutes
df_pivot = df.pivot_table(
values="number",
index="uids",
columns=["date_time"],
) #dataframe with all the uids on the index and all the datetimes in columns.
You can apply this to the whole dataframe or just a subset containing 6 rows. Then you rename your columns.

You can use the function crosstab:
df['uids'] = df['uids'].str.split(',')
df = df.explode('uids')
df['date_time'] = df['date_time'].dt.minute.floordiv(10).add(1)
pd.crosstab(df['uids'], df['date_time'], dropna=False)
Output:
date_time 1 2 3 4 5 6
uids
1000 1 0 0 2 0 0
1321 2 0 0 1 0 0
27 0 0 2 0 0 0
323 0 0 0 0 1 0
33 0 0 1 0 0 0
3974 0 0 2 0 2 0
691 0 0 0 0 1 0
7654 1 1 0 2 0 0
7665 0 0 1 0 0 0

We can achieve this with extracting the minutes from your datetime column. Then using pivot_table to get your wide format:
df['date_time'] = pd.to_datetime(df['date_time'])
df['minute'] = df['date_time'].dt.minute // 10
piv = (df.assign(uids=df['uids'].str.split(','))
.explode('uids')
.pivot_table(index='uids', columns='minute', values='minute', aggfunc='size')
)
minute 0 1 2 3 4
uids
1000 1.0 NaN NaN 2.0 NaN
1321 2.0 NaN NaN 1.0 NaN
27 NaN NaN 2.0 NaN NaN
323 NaN NaN NaN NaN 1.0
33 NaN NaN 1.0 NaN NaN
3974 NaN NaN 2.0 NaN 2.0
691 NaN NaN NaN NaN 1.0
7654 1.0 1.0 NaN 2.0 NaN
7665 NaN NaN 1.0 NaN NaN

How to find sum and count of a column based on a grouping condition on a Pandas dataset?

I have a Pandas dataset with 3 columns. I need to group by the ID column while finding the sum and count of the other two columns. Also, I have to ignore the zeroes in the columsn 'A' and 'B'.
The dataset looks like -
ID A B
1 0 5
2 10 0
2 20 0
3 0 30
What I need -
ID A_Count A_Sum B_Count B_Sum
1 0 0 1 5
2 2 30 0 0
3 0 0 1 30
I have tried this using one column but wasn't able to get both the aggregations in the final dataset.
(df.groupby('ID').agg({'A':'sum', 'A':'count'}).reset_index().rename(columns = {'A':'A_sum', 'A': 'A_count'}))

If you don't pass it columns specifically, it will aggregate the numeric columns by itself.
Since your don't want to count 0, replace them with NaN first:
df.replace(0, np.NaN, inplace=True)
print(df)
ID A B
0 1 NaN 5.0
1 2 10.0 NaN
2 2 20.0 NaN
3 3 NaN 30.0
df = df.groupby('ID').agg(['count', 'sum'])
print(df)
A B
count sum count sum
ID
1 0 0.0 1 5.0
2 2 30.0 0 0.0
3 0 0.0 1 30.0
Remove MultiIndex columns
You can use list comprehension:
df.columns = ['_'.join(col) for col in df.columns]
print(df)
A_count A_sum B_count B_sum
ID
1 0 0.0 1 5.0
2 2 30.0 0 0.0
3 0 0.0 1 30.0

Use Pandas dataframe to add lag feature from MultiIindex Series

I have a MultiIndex Series (3 indices) that looks like this:
Week ID_1 ID_2
3 26 1182 39.0
4767 42.0
31393 20.0
31690 42.0
32962 3.0
....................................
I also have a dataframe df which contains all the columns (and more) used for indices in the Series above, and I want to create a new column in my dataframe df that contains the value matching the ID_1 and ID_2 and the Week - 2 from the Series.
For example, for the row in dataframe that has ID_1 = 26, ID_2 = 1182 and Week = 3, I want to match the value in the Series indexed by ID_1 = 26, ID_2 = 1182 and Week = 1 (3-2) and put it on that row in a new column. Further, my Series might not necessarily have the value required by the dataframe, in which case I'd like to just have 0.
Right now, I am trying to do this by using:
[multiindex_series.get((x[1].get('week', 2) - 2, x[1].get('ID_1', 0), x[1].get('ID_2', 0))) for x in df.iterrows()]
This however is very slow and memory hungry and I was wondering what are some better ways to do this.
FWIW, the Series was created using
saved_groupby = df.groupby(['Week', 'ID_1', 'ID_2'])['Target'].median()
and I'm willing to do it a different way if better paths exist to create what I'm looking for.

Increase the Week by 2:
saved_groupby = df.groupby(['Week', 'ID_1', 'ID_2'])['Target'].median()
saved_groupby = saved_groupby.reset_index()
saved_groupby['Week'] = saved_groupby['Week'] + 2
and then merge df with saved_groupby:
result = pd.merge(df, saved_groupby, on=['Week', 'ID_1', 'ID_2'], how='left')
This will augment df with the target median from 2 weeks ago.
To make the median (target) saved_groupby column 0 when there is no match, use fillna to change NaNs to 0:
result['Median'] = result['Median'].fillna(0)
For example,
import numpy as np
import pandas as pd
np.random.seed(2016)
df = pd.DataFrame(np.random.randint(5, size=(20,5)),
columns=['Week', 'ID_1', 'ID_2', 'Target', 'Foo'])
saved_groupby = df.groupby(['Week', 'ID_1', 'ID_2'])['Target'].median()
saved_groupby = saved_groupby.reset_index()
saved_groupby['Week'] = saved_groupby['Week'] + 2
saved_groupby = saved_groupby.rename(columns={'Target':'Median'})
result = pd.merge(df, saved_groupby, on=['Week', 'ID_1', 'ID_2'], how='left')
result['Median'] = result['Median'].fillna(0)
print(result)
yields
Week ID_1 ID_2 Target Foo Median
0 3 2 3 4 2 0.0
1 3 3 0 3 4 0.0
2 4 3 0 1 2 0.0
3 3 4 1 1 1 0.0
4 2 4 2 0 3 2.0
5 1 0 1 4 4 0.0
6 2 3 4 0 0 0.0
7 4 0 0 2 3 0.0
8 3 4 3 2 2 0.0
9 2 2 4 0 1 0.0
10 2 0 4 4 2 0.0
11 1 1 3 0 0 0.0
12 0 1 0 2 0 0.0
13 4 0 4 0 3 4.0
14 1 2 1 3 1 0.0
15 3 0 1 3 4 2.0
16 0 4 2 2 4 0.0
17 1 1 4 4 2 0.0
18 4 1 0 3 0 0.0
19 1 0 1 0 0 0.0