Can we replace outliers with the predicted values in pyspark? - apache-spark
I have a df in spark:
(I am actually working on this dataset it is not possible to paste whole data so here is the link)
df = https://www.kaggle.com/schirmerchad/bostonhoustingmlnd?select=housing.csv
Now I found the outliers as below (22 rows in total):
def IQR(df,column):
quantiles = sdf.approxQuantile(column, [0.25, 0.75], 0)
q1 = quantiles[0]
q3 = quantiles[1]
IQR = q3-q1
lower = q1 - 1.5*IQR
upper = q3+ 1.5*IQR
return (lower,upper)
lower, upper = IQR(df,'RM')
lower,upper = 4.8374999999999995 7.617500000000001
outliers = df.filter((df['RM'] > upper) | (df['RM'] < lower))
Now below are the outliers detected :
RM LSTAT PTRATIO MEDV
8.069 4.21 18 812700
7.82 3.57 18 919800
7.765 7.56 17.8 835800
7.853 3.81 14.7 1018500
8.266 4.14 17.4 940800
8.04 3.13 17.4 789600
7.686 3.92 17.4 980700
8.337 2.47 17.4 875700
8.247 3.95 17.4 1014300
8.259 3.54 19.1 898800
8.398 5.91 13 1024800
7.691 6.58 18.6 739200
7.82 3.76 14.9 953400
7.645 3.01 14.9 966000
3.561 7.12 20.2 577500
3.863 13.33 20.2 485100
4.138 37.97 20.2 289800
4.368 30.63 20.2 184800
4.652 28.28 20.2 220500
4.138 23.34 20.2 249900
4.628 34.37 20.2 375900
4.519 36.98 20.2 147000
Now I want to replace the outliers with the ml predicted values, after the ml process I got the predicted values as below:-
RM LSTAT PTRATIO MEDV column_assem column prediction
8.069 4.21 18 812700 {"vectorType":"dense","length":3,"values":[4.21,18,812700]} {"vectorType":"dense","length":3,"values":[812699.9991344779,32.9872628621034,25.697942748362507]} 7.138307692307692
7.82 3.57 18 919800 {"vectorType":"dense","length":3,"values":[3.57,18,919800]} {"vectorType":"dense","length":3,"values":[919799.999082192,36.25675952004636,26.656936598060938]} 7.138307692307692
7.765 7.56 17.8 835800 {"vectorType":"dense","length":3,"values":[7.56,17.8,835800]} {"vectorType":"dense","length":3,"values":[835799.9989959698,37.18609141885786,25.87518521779868]} 7.138307692307692
7.853 3.81 14.7 1018500 {"vectorType":"dense","length":3,"values":[3.81,14.7,1018500]} {"vectorType":"dense","length":3,"values":[1018499.9990279829,40.25963007114179,24.285126110831364]} 7.138307692307692
8.266 4.14 17.4 940800 {"vectorType":"dense","length":3,"values":[4.14,17.4,940800]} {"vectorType":"dense","length":3,"values":[940799.9990507461,37.621770135316275,26.279618209844216]} 7.138307692307692
8.04 3.13 17.4 789600 {"vectorType":"dense","length":3,"values":[3.13,17.4,789600]} {"vectorType":"dense","length":3,"values":[789599.999195178,31.094759131505864,24.832393813608636]} 7.138307692307692
7.686 3.92 17.4 980700 {"vectorType":"dense","length":3,"values":[3.92,17.4,980700]} {"vectorType":"dense","length":3,"values":[980699.9990305867,38.858227336579965,26.637789595102927]} 7.138307692307692
8.337 2.47 17.4 875700 {"vectorType":"dense","length":3,"values":[2.47,17.4,875700]} {"vectorType":"dense","length":3,"values":[875699.9991585133,33.577861049146954,25.59625197564997]} 7.138307692307692
8.247 3.95 17.4 1014300 {"vectorType":"dense","length":3,"values":[3.95,17.4,1014300]} {"vectorType":"dense","length":3,"values":[1014299.9990056665,40.11446130241714,26.949909126197]} 7.138307692307692
8.259 3.54 19.1 898800 {"vectorType":"dense","length":3,"values":[3.54,19.1,898800]} {"vectorType":"dense","length":3,"values":[898799.9990899825,35.406713649671325,27.56000332051734]} 7.138307692307692
8.398 5.91 13 1024800 {"vectorType":"dense","length":3,"values":[5.91,13,1024800]} {"vectorType":"dense","length":3,"values":[1024799.9989586923,42.669988999612016,22.74784587477886]} 7.138307692307692
7.691 6.58 18.6 739200 {"vectorType":"dense","length":3,"values":[6.58,18.6,739200]} {"vectorType":"dense","length":3,"values":[739199.9990946348,32.64270527156902,25.73328780757773]} 7.138307692307692
7.82 3.76 14.9 953400 {"vectorType":"dense","length":3,"values":[3.76,14.9,953400]} {"vectorType":"dense","length":3,"values":[953399.9990744753,37.82403517229104,23.880552758747136]} 7.138307692307692
7.645 3.01 14.9 966000 {"vectorType":"dense","length":3,"values":[3.01,14.9,966000]} {"vectorType":"dense","length":3,"values":[965999.9990932231,37.53477931241747,23.960460322415766]} 7.138307692307692
3.561 7.12 20.2 577500 {"vectorType":"dense","length":3,"values":[7.12,20.2,577500]} {"vectorType":"dense","length":3,"values":[577499.9991773808,27.20258411502299,25.862694427868608]} 6.376732394366198
3.863 13.33 20.2 485100 {"vectorType":"dense","length":3,"values":[13.33,20.2,485100]} {"vectorType":"dense","length":3,"values":[485099.999013695,30.032948373359417,25.311342678468208]} 6.043858108108108
4.138 37.97 20.2 289800 {"vectorType":"dense","length":3,"values":[37.97,20.2,289800]} {"vectorType":"dense","length":3,"values":[289799.99824280146,47.51591753902686,24.707706732637366]} 5.2370714285714275
4.368 30.63 20.2 184800 {"vectorType":"dense","length":3,"values":[30.63,20.2,184800]} {"vectorType":"dense","length":3,"values":[184799.99858809082,36.35256433967503,23.378827944979733]} 5.2370714285714275
4.652 28.28 20.2 220500 {"vectorType":"dense","length":3,"values":[28.28,20.2,220500]} {"vectorType":"dense","length":3,"values":[220499.9986495131,35.3082739723793,23.59425617851294]} 5.2370714285714275
4.138 23.34 20.2 249900 {"vectorType":"dense","length":3,"values":[23.34,20.2,249900]} {"vectorType":"dense","length":3,"values":[249899.99881098093,31.44714189260281,23.625084354536643]} 6.043858108108108
4.628 34.37 20.2 375900 {"vectorType":"dense","length":3,"values":[34.37,20.2,375900]} {"vectorType":"dense","length":3,"values":[375899.9983146336,47.06252004732307,25.328138233469573]} 5.2370714285714275
4.519 36.98 20.2 147000 {"vectorType":"dense","length":3,"values":[36.98,20.2,147000]} {"vectorType":"dense","length":3,"values":[146999.99838054206,41.31545014321207,23.33912202640834]} 5.2370714285714275
If it is one value I am aware of lit() to replace it but when there are multiple values how do we replace with the original one's?
Assuming that the original dataframe is called df and the machine-learning transformed dataframe is called ml, you can do a join and replace the RM column with the prediction value if the row satisfy the outlier condition:
df2 = df.join(ml, df.columns, 'left').withColumn(
'RM',
F.when(
(F.col('RM') > upper) | (F.col('RM') < lower),
F.col('prediction')
).otherwise(F.col('RM'))
).select(df.columns)
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It is possible to set the precision in the following way: for i in range(100): k = 1 + i / 100; print("%.Nf"%k) Where N - decimal numbers. Keep in mind, that regularly you don't need a lot of them, though the number could be really huge.
Extending macro from 1 row to 56 rows. Application defined error
Ihave never done Excel VBA macros. The data I’m trying to get organized into a single column is in excel rows 22-78. 0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48 0.52 0.56 0.6 0.64 0.68 0.72 0.76 0.8 0.84 0.88 0.92 0.96 1 1.04 1.08 1.12 1.16 1.2 1.24 1.28 1.32 1.36 1.4 1.44 1.48 1.52 1.56 1.6 1.64 1.68 1.72 1.76 1.8 1.84 1.88 1.92 1.96 2 2.04 2.08 2.12 2.16 2.2 2.24 2.28 2.32 2.36 2.4 2.44 2.48 2.52 2.56 2.6 2.64 2.68 2.72 2.76 2.8 2.84 2.88 2.92 2.96 3 3.04 3.08 3.12 3.16 3.2 3.24 3.28 3.32 3.36 3.4 3.44 3.48 3.52 3.56 3.6 3.64 3.68 3.72 3.76 3.8 3.84 3.88 3.92 3.96 4 4.04 4.08 4.12 4.16 4.2 4.24 4.28 4.32 4.36 4.4 4.44 4.48 4.52 4.56 4.6 4.64 4.68 4.72 4.76 4.8 4.84 4.88 4.92 4.96 5 5.04 5.08 5.12 5.16 5.2 5.24 5.28 5.32 5.36 5.4 5.44 5.48 5.52 5.56 5.6 5.64 5.68 5.72 5.76 5.8 5.84 5.88 5.92 5.96 6 6.04 6.08 6.12 6.16 6.2 6.24 6.28 6.32 6.36 6.4 6.44 6.48 6.52 6.56 6.6 6.64 6.68 6.72 6.76 6.8 6.84 6.88 6.92 6.96 7 7.04 7.08 7.12 7.16 7.2 7.24 7.28 7.32 7.36 7.4 7.44 7.48 7.52 7.56 7.6 7.64 7.68 7.72 7.76 7.8 7.84 7.88 7.92 7.96 8 8.04 8.08 8.12 8.16 8.2 8.24 8.28 8.32 8.36 8.4 8.44 8.48 8.52 8.56 8.6 8.64 8.68 8.72 8.76 8.8 8.84 8.88 8.92 8.96 9 9.04 9.08 9.12 9.16 9.2 9.24 9.28 9.32 9.36 9.4 9.44 9.48 9.52 9.56 9.6 9.64 9.68 9.72 9.76 9.8 9.84 9.88 9.92 9.96 10 10.04 10.08 10.12 10.16 10.2 10.24 10.28 10.32 10.36 10.4 10.44 10.48 10.52 10.56 10.6 10.64 10.68 10.72 10.76 10.8 10.84 10.88 10.92 10.96 11 11.04 11.08 11.12 11.16 11.2 11.24 11.28 11.32 11.36 11.4 11.44 11.48 11.52 11.56 11.6 11.64 11.68 11.72 11.76 11.8 11.84 11.88 11.92 11.96 12 12.04 12.08 12.12 12.16 12.2 12.24 12.28 12.32 12.36 12.4 12.44 12.48 12.52 12.56 12.6 12.64 12.68 12.72 12.76 12.8 12.84 12.88 12.92 12.96 13 13.04 13.08 13.12 13.16 13.2 13.24 13.28 13.32 13.36 13.4 13.44 13.48 13.52 13.56 13.6 13.64 13.68 13.72 13.76 13.8 13.84 13.88 13.92 13.96 14 14.04 14.08 14.12 14.16 14.2 14.24 14.28 14.32 14.36 14.4 14.44 14.48 14.52 14.56 14.6 14.64 14.68 14.72 14.76 14.8 14.84 14.88 14.92 14.96 15 15.04 15.08 15.12 15.16 15.2 15.24 15.28 15.32 This is the data in one row. And such I have from row 22-78. the final files have a similar number of columns but many more rows. I am not sure what would be a good way to organize this into a single column in excel I got this working for 1 row. here's the code Sub RowsToColumn() Dim RN As Range Dim RI As Range Dim r As Long Dim LR As Long Application.ScreenUpdating = False Columns(1).Insert r = 0 LR = Range("A" & Rows.Count).End(xlUp).row For Each RN In Range("A1:A" & LR) r = r + 1 For Each RI In Range(RN, Range("XFD" & RN.row).End(xlToLeft)) r = r + 1 Cells(r, 1) = RI RI.Clear Next RI Next RN Columns("A:A").SpecialCells(xlCellTypeBlanks).Delete Shift:=xlUp End Sub But to extend this for Rows A22-78 Sub RowsToColumn_Second() Dim RN As Range Dim RI As Range Dim r As Long Dim LR As Long Dim row As Range Dim rng As Range Dim cell As Range Application.ScreenUpdating = False Set rng = Range("A22:A78") For Each row In rng.Rows Columns(1, rng).Insert r = 0 LR = Range("A" & Rows.Count).End(xlUp).row LR = Range("A" & Rows.Count).End(xlUp).row For Each RN In Range("A1:A" & LR) r = r + 1 For Each RI In Range(RN, Range("XFD" & RN.row).End(xlToLeft)) r = r + 1 Cells(r, 1) = RI RI.Clear Next RI Next RN Next row Columns("A:A").SpecialCells(xlCellTypeBlanks).Delete Shift:=xlUp End Sub This is where it saysApplication defined error-1004. It doesn't like Columns(1, rng).Insert
copy the data and Paste Special -> Transpose, this will change from rows to colums, or viceversa