pyspark - assign non-null columns to new columns - apache-spark

I have a dataframe of the following scheme in pyspark:
user_id datadate page_1.A page_1.B page_1.C page_2.A page_2.B \
0 111 20220203 NaN NaN NaN NaN NaN
1 222 20220203 5 5 5 5.0 5.0
2 333 20220203 3 3 3 3.0 3.0
page_2.C page_3.A page_3.B page_3.C
0 NaN 1.0 1.0 2.0
1 5.0 NaN NaN NaN
2 4.0 NaN NaN NaN
So it contains columns like user_id, datadate, and few columns for each page (got 3 pages), which are the result of 2 joins. In this example, i have page_1, page_2, page_3, and each has 3 columns: A,B,C. Additionally, for each page columns, for each row, they will either be all null or all full, like in my example.
I don't care about the values of each of the columns per page, I just want to get for each row, the [A,B,C] values that are not null.
example for a wanted result table:
user_id datadate A B C
0 111 20220203 1 1 2
1 222 20220203 5 5 5
2 333 20220203 3 3 3
so the logic will be something like:
df[A] = page_1.A or page_2.A or page_3.A, whichever is not null
df[B] = page_1.B or page_2.B or page_3.B, whichever is not null
df[C] = page_1.C or page_2.C or page_3.C, whichever is not null
for all of the rows..
and of course, I would like to do it in an efficient way.
Thanks a lot.

You can use the sql functions greatest to extract the greatest values in a list of columns.
You can find the documentation here: https://spark.apache.org/docs/3.1.1/api/python/reference/api/pyspark.sql.functions.greatest.html
from pyspark.sql import functions as F
(df.withColumn('A', F.greates(F.col('page_1.A'), F.col('page_2.A), F.col('page_3.A'))
.withColumn('B', F.greates(F.col('page_1.B'), F.col('page_2.B), F.col('page_3.B'))
.select('userid', 'datadate', 'A', 'B'))

Related

Join with column having the max sequence number

I have a margin table
item margin
0 a 3
1 b 4
2 c 5
and an item table
item sequence
0 a 1
1 a 2
2 a 3
3 b 1
4 b 2
5 c 1
6 c 2
7 c 3
I want to join the two table so that the margin will only be joined to the product with maximum sequence number, the desired outcome is
item sequence margin
0 a 1 NaN
1 a 2 NaN
2 a 3 3.0
3 b 1 NaN
4 b 2 4.0
5 c 1 NaN
6 c 2 NaN
7 c 3 5.0
How to achieve this?
Below is the code for margin and item table
import pandas as pd
df_margin=pd.DataFrame({"item":["a","b","c"],"margin":[3,4,5]})
df_item=pd.DataFrame({"item":["a","a","a","b","b","c","c","c"],"sequence":[1,2,3,1,2,1,2,3]})
One option would be to merge then replace extra values with NaN via Series.where:
new_df = df_item.merge(df_margin)
new_df['margin'] = new_df['margin'].where(
new_df.groupby('item')['sequence'].transform('max').eq(new_df['sequence'])
)
Or with loc:
new_df = df_item.merge(df_margin)
new_df.loc[new_df.groupby('item')['sequence']
.transform('max').ne(new_df['sequence']), 'margin'] = np.NAN
Another option would be to assign a temp column to both frames df_item with True where the value is maximal, and df_margin is True everywhere then merge outer and drop the temp column:
new_df = (
df_item.assign(
t=df_item
.groupby('item')['sequence']
.transform('max')
.eq(df_item['sequence'])
).merge(df_margin.assign(t=True), how='outer').drop('t', 1)
)
Both produce new_df:
item sequence margin
0 a 1 NaN
1 a 2 NaN
2 a 3 3.0
3 b 1 NaN
4 b 2 4.0
5 c 1 NaN
6 c 2 NaN
7 c 3 5.0
You could do:
df_item.merge(df_item.groupby('item')['sequence'].max().\
reset_index().merge(df_margin), 'left')
item sequence margin
0 a 1 NaN
1 a 2 NaN
2 a 3 3.0
3 b 1 NaN
4 b 2 4.0
5 c 1 NaN
6 c 2 NaN
7 c 3 5.0
Breakdown:
df_new = df_item.groupby('item')['sequence'].max().reset_index().merge(df_margin)
df_item.merge(df_new, 'left')

pandas groupby and widen dataframe with ordered columns

I have a long form dataframe that contains multiple samples and time points for each subject. The number of samples and timepoint can vary, and the days between time points can also vary:
test_df = pd.DataFrame({"subject_id":[1,1,1,2,2,3],
"sample":["A", "B", "C", "D", "E", "F"],
"timepoint":[19,11,8,6,2,12],
"time_order":[3,2,1,2,1,1]
})
subject_id sample timepoint time_order
0 1 A 19 3
1 1 B 11 2
2 1 C 8 1
3 2 D 6 2
4 2 E 2 1
5 3 F 12 1
I need to figure out a way to generalize grouping this dataframe by subject_id and putting all samples and time points on the same row, in time order.
DESIRED OUTPUT:
subject_id sample1 timepoint1 sample2 timepoint2 sample3 timepoint3
0 1 C 8 B 11 A 19
1 2 E 2 D 6 null null
5 3 F 12 null null null null
Pivot gets me close, but I'm stuck on how to proceed from there:
test_df = test_df.pivot(index=['subject_id', 'sample'],
columns='time_order', values='timepoint')
Use DataFrame.set_index with DataFrame.unstack for pivoting, sorting MultiIndex in columns, flatten it and last convert subject_id to column:
df = (test_df.set_index(['subject_id', 'time_order'])
.unstack()
.sort_index(level=[1,0], axis=1))
df.columns = df.columns.map(lambda x: f'{x[0]}{x[1]}')
df = df.reset_index()
print (df)
subject_id sample1 timepoint1 sample2 timepoint2 sample3 timepoint3
0 1 C 8.0 B 11.0 A 19.0
1 2 E 2.0 D 6.0 NaN NaN
2 3 F 12.0 NaN NaN NaN NaN
a=test_df.iloc[:,:3].groupby('subject_id').last().add_suffix('1')
b=test_df.iloc[:,:3].groupby('subject_id').nth(-2).add_suffix('2')
c=test_df.iloc[:,:3].groupby('subject_id').nth(-3).add_suffix('3')
pd.concat([a, b,c], axis=1)
sample1 timepoint1 sample2 timepoint2 sample3 timepoint3
subject_id
1 C 8 B 11.0 A 19.0
2 E 2 D 6.0 NaN NaN
3 F 12 NaN NaN NaN NaN

Create Python function to look for ANY NULL value in a group

I am trying to write a function that will check a specified column for nulls, within a group in a dataframe. The example dataframe has two columns, ID and VALUE. Multiple rows exist per ID. I want to know if ANY of the rows for a particular ID have a NULL value in VALUE.
I have tried building the function with iterrows().
df = pd.DataFrame({'ID':[1,2,2,3,3,3],
'VALUE':[50,None,30,20,10,None]})
def nullValue(col):
for i, row in col.iterrows():
if ['VALUE'] is None:
return 1
else:
return 0
df2 = df.groupby('ID').apply(nullVALUE)
df2.columns = ['ID','VALUE','isNULL']
df2
I am expecting to retrieve a dataframe with three columns, ID, VALUE, and isNULL. If any row in a grouped ID has a null, all of the rows for that ID should have a 1 under isNull.
Example:
ID VALUE isNULL
1 50.0 0
2 NaN 1
2 30.0 1
3 20.0 1
3 10.0 1
3 NaN 1
A quick solution, borrowed partially from this answer is to use groupby with transform:
df = pd.DataFrame({'ID':[1,2,2,3,3,3,3],
'VALUE':[50,None,None,30,20,10,None]})
df['isNULL'] = (df.VALUE.isnull().groupby([df['ID']]).transform('sum') > 0).astype(int)
Out[51]:
ID VALUE isNULL
0 1 50.0 0
1 2 NaN 1
2 2 NaN 1
3 3 30.0 1
4 3 20.0 1
5 3 10.0 1
6 3 NaN 1

Pandas append returns DF with NaN values

I'm appending data from a list to pandas df. I keep getting NaN in my entries.
Based on what I've read I think I might have to mention the data type for each column in my code.
dumps = [];features_df = pd.DataFrame()
for i in range (int(len(ids)/50)):
dumps = sp.audio_features(ids[i*50:50*(i+1)])
for i in range (len(dumps)):
print(list(dumps[0].values()))
features_df = features_df.append(list(dumps[0].values()), ignore_index = True)
Expected results, something like-
[0.833, 0.539, 11, -7.399, 0, 0.178, 0.163, 2.1e-06, 0.101, 0.385, 99.947, 'audio_features', '6MWtB6iiXyIwun0YzU6DFP', 'spotify:track:6MWtB6iiXyIwun0YzU6DFP', 'https://api.spotify.com/v1/tracks/6MWtB6iiXyIwun0YzU6DFP', 'https://api.spotify.com/v1/audio-analysis/6MWtB6iiXyIwun0YzU6DFP', 149520, 4]
for one row.
Actual-
danceability energy ... duration_ms time_signature
0 NaN NaN ... NaN NaN
1 NaN NaN ... NaN NaN
2 NaN NaN ... NaN NaN
3 NaN NaN ... NaN NaN
4 NaN NaN ... NaN NaN
5 NaN NaN ... NaN NaN
For all rows
append() strategy in a tight loop isn't a great way to do this. Rather, you can construct an empty DataFrame and then use loc to specify an insertion point. The DataFrame index should be used.
For example:
import pandas as pd
df = pd.DataFrame(data=[], columns=['n'])
for i in range(100):
df.loc[i] = i
print(df)
time python3 append_df.py
n
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
real 0m13.178s
user 0m12.287s
sys 0m0.617s
https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.append.html
Iteratively appending rows to a DataFrame can be more computationally intensive than a single concatenate. A better solution is to append those rows to a list and then concatenate the list with the original DataFrame all at once.

Replacing values in specific columns in a Pandas Dataframe, when number of columns are unknown

I am brand new to Python and stacks exchange. I have been trying to replace invalid values ( x<-3 and x>12) with np.nan in specific columns.
I don't know how many columns I will have to deal with and thus will have to create a general code that takes this into account. I do however know, that the first two columns are ids and names respectively. I have searched google and stacks exchange for a solution but haven't been able to find a solution that solves my specific objective.
My question is; How would one replace values found in the third column and onwards?
My dataframe looks like this;
Data
I tried this line:
Data[Data > 12.0] = np.nan.
this replaced the first two columns with nan
1st attempt
I tried this line:
Data[(Data.iloc[(range(2,Columns))] >=12) & (Data.iloc[(range(2,Columns))]<=-3)] = np.nan
where,
Columns = len(Data.columns)
This is clearly wrong replacing all values in rows 2 to 6 (Columns = 7).
2nd attempt
Any thoughts would be greatly appreciated.
Python 3.6.1 64bits, Qt 5.6.2, PyQt5 5.6 on Darwin
You're looking for the applymap() method.
import pandas as pd
import numpy as np
# get the columns after the second one
cols = Data.columns[2:]
# apply mask to those columns
new_df = Data[cols].applymap(lambda x: np.nan if x > 12 or x <= -3 else x)
Documentation: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.applymap.html
This approach assumes your columns after the second contain float or int values.
You can set values to specific columns of a dataframe by using iloc and slicing the columns that you need. Then we can set the values using where
A short example using some random data
df = pd.DataFrame(np.random.randint(0,10,(4,10)))
0 1 2 3 4 5 6 7 8 9
0 7 7 9 4 2 6 6 1 7 9
1 0 1 2 4 5 5 3 9 0 7
2 0 1 4 4 3 8 7 0 6 1
3 1 4 0 2 5 7 2 7 9 9
Now we set the region to update and the region we want to update using iloc, and we slice columns indexed as 2 to the last column
df.iloc[:,2:] = df.iloc[:,2:].where((df < 7) & (df > 2))
Which will set the values in the Data Frame to NaN.
0 1 2 3 4 5 6 7 8 9
0 7 7 NaN 4.0 NaN 6.0 6.0 NaN NaN NaN
1 0 1 NaN 4.0 5.0 5.0 3.0 NaN NaN NaN
2 0 1 4.0 4.0 3.0 NaN NaN NaN 6.0 NaN
3 1 4 NaN NaN 5.0 NaN NaN NaN NaN NaN
For your data the code would be this
Data.iloc[:,2:] = Data.iloc[:,2:].where((Data <= 12) & (Data >= -3))
Operator clarification
The setup I show directly above would look like this
-3 <= Data <= 12, gives everything between those numbers
If we reverse this logic using the & operator it looks like this
-3 >= Data <= 12, a number cannot be both less than -3 and greater than 12 at the same time.
So we use the or operator instead |. Code looks like this now....
Data.iloc[:,2:] = Data.iloc[:,2:].where((Data >= 12) | (Data <= -3))
So the data is checked on a conditional basis
Data <= -3 or Data >= 12

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