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How to combine two pandas dataset based on multiple conditions?

I want to combine two datasets in Python based on multiple conditions using pandas.
The two datasets are different numbers of rows.
The first one contains almost 300k entries, while the second one contains almost 1000 entries.
More specifically, The first dataset: "A" has the following information:
Path | Line | Severity | Vulnerability | Name | Text | Title
An instance of the content of "A" is this:
src.bla.bla.class.java| 24; medium| Logging found| hr.kravarscan.enchantedfortress_15| description| Enchanted Fortress
While the second dataset: "B" contains the following information:
Class | Path | DTWC | DR | DW | IDFP
An instance of the content in "B" is this:
y.x.bla.MainActivity | com.lucao.limpazap_11| 0 | 0 | 0 | 0
I want to combine these two dataset as follow:
If A['Name'] is equal to B['Path'] AND B['Class'] is in A['Class']
Than
Merge the two lines into another data frame "C"
An output example is the following:
Suppose that A contains:
src.bla.bla.class.java| 24| medium| Logging found| hr.kravarscan.enchantedfortress_15| description| Enchanted Fortress|
and B contains:
com.bla.class | hr.kravarscan.enchantedfortress_15| 0 | 0 | 0 | 0
the output should be the following:
src.bla.bla.class.java| 24| medium| Logging found| hr.kravarscan.enchantedfortress_15| description| Enchanted Fortress| com.bla.class | hr.kravarscan.enchantedfortress_15| 0 | 0 | 0 | 0

I'm not sure if this the best and the most efficient way but i have test it and it worked. So my answer is pretty straight forward, we will loop over two dataframes and apply the desired conditions.
Suppose the dataset A is df_a and dataset B is df_b.
First we have to add a suffix on every columns on df_a and df_b so both rows can be appended later.
df_a.columns= [i+'_A' for i in df_a.columns]
df_b.columns= [i+'_B' for i in df_b.columns]
And then we can apply this for loop
df_c= pd.DataFrame()
# Iterate through df_a
for (idx_A, v_A) in df_a.iterrows():
# Iterate through df_b
for (idx_B, v_B) in df_b.iterrows():
# Apply the condition
if v_A['Name_A']==v_B['Path_B'] and v_B['Class_B'] in v_A['Path_A']:
# Cast both series to dictionary and then append them to a new dict
c_dict= {**v_A.to_dict(), **v_B.to_dict()}
# Append the df_c with c_dict
df_c= df_c.append(c_dict, ignore_index=True)

create a column that is the sum of previous X rows where x is a parm given by a different column row

Im trying to create a column where i sum the previous x rows of a column by a parm given in a different column row.
I have a solution but its really slow so i was wondering if anyone could help do this alot faster.
| time | price |parm |
|--------------------------|------------|-----|
|2020-11-04 00:00:00+00:00 | 1.17600 | 1 |
|2020-11-04 00:01:00+00:00 | 1.17503 | 2 |
|2020-11-04 00:02:00+00:00 | 1.17341 | 3 |
|2020-11-04 00:03:00+00:00 | 1.17352 | 2 |
|2020-11-04 00:04:00+00:00 | 1.17422 | 3 |
and the slow slow code
#jit
def rolling_sum(x,w):
return np.convolve(x,np.ones(w,dtype=int),'valid')
#jit
def rol(x,y):
for i in range(len(x)):
res[i] = rolling_sum(x, y[i])[0]
return res
dfa = df[:500000]
res = np.empty(len(dfa))
r = rol(dfa.l_x.values, abs(dfa.mb).values+1)
r

Maybe something like this could work. I have made up an example with to_be_summed being the column of the value that should be summed up and looback holding the number of rows to be looked back
df = pd.DataFrame({"to_be_summed": range(10), "lookback":[0,1,2,3,2,1,4,2,1,2]})
summed = df.to_be_summed.cumsum()
result = [summed[i] - summed[max(0,i - lookback - 1)] for i, lookback in enumerate(df.lookback)]
What I did here is to first do a cumsum over the column that should be summed up. Now, for the i-th entry I can take the entry of this cumsum, and subtract the one i + 1 steps back. Note that this include the i-th value in the sum. If you don't want to inlcude it, you just have to change from summed[i] to summed[i - 1]. Also note that this part max(0,i - lookback - 1) will prevent you from accidentally looking back too many rows.

Get Index Minimum Value in Column When String - Pandas Dataframe

I have done some research on this, but couldn't find a concise method when the index is of type 'string'.
Given the following Pandas dataframe:
Platform | Action | RPG | Fighting
----------------------------------------
PC | 4 | 6 | 9
Playstat | 6 | 7 | 5
Xbox | 9 | 4 | 6
Wii | 8 | 8 | 7
I was trying to get the index (Platform) of the smallest value in the 'RPG' column, which would return 'Xbox'. I managed to make it work but it's not efficient, and looking for a better/quicker/condensed approach. Here is what I got:
# Return the minimum value of a series of all columns values for RPG
series1 = min(ign_data.loc['RPG'])
# Find the lowest value in the series
minim = min(ign_data.loc['RPG'])
# Get the index of that value using boolean indexing
result = series1[series1 == minim].index
# Format that index to a list, and return the first (and only) element
str_result = result.format()[0]

Use Series.idxmin:
df.set_index('Platform')['RPG'].idxmin()
#'Xbox'
or what #Quang Hoang suggests on the comments
df.loc[df['RPG'].idxmin(), 'Platform']
if Platform already the index:
df['RPG'].idxmin()
EDIT
df.set_index('Platform').loc['Playstat'].idxmin()
#'Fighting'
df.set_index('Platform').idxmin(axis=1)['Playstat']
#'Fighting'
if already the index:
df.loc['Playstat'].idxmin()

How to add new row in pandas dataframe? [duplicate]

I have an existing dataframe which I need to add an additional column to which will contain the same value for every row.
Existing df:
Date, Open, High, Low, Close
01-01-2015, 565, 600, 400, 450
New df:
Name, Date, Open, High, Low, Close
abc, 01-01-2015, 565, 600, 400, 450
I know how to append an existing series / dataframe column. But this is a different situation, because all I need is to add the 'Name' column and set every row to the same value, in this case 'abc'.

df['Name']='abc' will add the new column and set all rows to that value:
In [79]:
df
Out[79]:
Date, Open, High, Low, Close
0 01-01-2015, 565, 600, 400, 450
In [80]:
df['Name'] = 'abc'
df
Out[80]:
Date, Open, High, Low, Close Name
0 01-01-2015, 565, 600, 400, 450 abc

You can use insert to specify where you want to new column to be. In this case, I use 0 to place the new column at the left.
df.insert(0, 'Name', 'abc')
Name Date Open High Low Close
0 abc 01-01-2015 565 600 400 450

Summing up what the others have suggested, and adding a third way
You can:
assign(**kwargs):
df.assign(Name='abc')
access the new column series (it will be created) and set it:
df['Name'] = 'abc'
insert(loc, column, value, allow_duplicates=False)
df.insert(0, 'Name', 'abc')
where the argument loc ( 0 <= loc <= len(columns) ) allows you to insert the column where you want.
'loc' gives you the index that your column will be at after the insertion. For example, the code above inserts the column Name as the 0-th column, i.e. it will be inserted before the first column, becoming the new first column. (Indexing starts from 0).
All these methods allow you to add a new column from a Series as well (just substitute the 'abc' default argument above with the series).

Single liner works
df['Name'] = 'abc'
Creates a Name column and sets all rows to abc value

I want to draw more attention to a portion of #michele-piccolini's answer.
I strongly believe that .assign is the best solution here. In the real world, these operations are not in isolation, but in a chain of operations. And if you want to support a chain of operations, you should probably use the .assign method.
Here is an example using snowfall data at a ski resort (but the same principles would apply to say ... financial data).
This code reads like a recipe of steps. Both assignment (with =) and .insert make this much harder:
raw = pd.read_csv('https://github.com/mattharrison/datasets/raw/master/data/alta-noaa-1980-2019.csv',
parse_dates=['DATE'])
def clean_alta(df):
return (df
.loc[:, ['STATION', 'NAME', 'LATITUDE', 'LONGITUDE', 'ELEVATION', 'DATE',
'PRCP', 'SNOW', 'SNWD', 'TMAX', 'TMIN', 'TOBS']]
.groupby(pd.Grouper(key='DATE', freq='W'))
.agg({'PRCP': 'sum', 'TMAX': 'max', 'TMIN': 'min', 'SNOW': 'sum', 'SNWD': 'mean'})
.assign(LOCATION='Alta',
T_RANGE=lambda w_df: w_df.TMAX-w_df.TMIN)
)
clean_alta(raw)
Notice the line .assign(LOCATION='Alta', that creates a column with a single value in the middle of the rest of the operations.

One Line did the job for me.
df['New Column'] = 'Constant Value'
df['New Column'] = 123

You can Simply do the following:
df['New Col'] = pd.Series(["abc" for x in range(len(df.index))])

This single line will work.
df['name'] = 'abc'

The append method has been deprecated since Pandas 1.4.0
So instead use the above method only if using actual pandas DataFrame object:
df["column"] = "value"
Or, if setting value on a view of a copy of a DataFrame, use concat() or assign():
This way the new Series created has the same index as original DataFrame, and so will match on exact rows
# adds a new column in view `where_there_is_one` named
# `client` with value `display_name`
# `df` remains unchanged
df = pd.DataFrame({"number": ([1]*5 + [0]*5 )})
where_there_is_one = df[ df["number"] == 1]
where_there_is_one = pd.concat([
where_there_is_one,
pd.Series(["display_name"]*df.shape[0],
index=df.index,
name="client")
],
join="inner", axis=1)
# Or use assign
where_there_is_one = where_there_is_one.assign(client = "display_name")
Output:
where_there_is_one: df:
| 0 | number | client | | 0 | number |
| --- | --- | --- | |---| -------|
| 0 | 1 | display_name | | 0 | 1 |
| 1 | 1 | display_name | | 1 | 1 |
| 2 | 1 | display_name | | 2 | 1 |
| 3 | 1 | display_name | | 3 | 1 |
| 4 | 1 | display_name | | 4 | 1 |
| 5 | 0 |
| 6 | 0 |
| 7 | 0 |
| 8 | 0 |
| 9 | 0 |

Ok, all, I have a similar situation here but if i take this code to use: df['Name']='abc'
instead 'abc' the name for the new column I want to take from somewhere else in the csv file.
As you can see from the picture, df is not cleaned yet but I want to create 2 columns with the name "ADI dms rivoli" which will continue for every row, and the same for the "December 2019". Hope it is clear for you to understand, it was hard to explaine, sorry.

Merge two dataframes in PySpark

I have two dataframes, DF1 and DF2, DF1 is the master which stores any additional information from DF2.
Lets say the DF1 is of the following format,
Item Id | item | count
---------------------------
1 | item 1 | 2
2 | item 2 | 3
1 | item 3 | 2
3 | item 4 | 5
DF2 contains the 2 items which were already present in DF1 and two new entries. (itemId and item are considered as a single group, can be treated as the key for join)
Item Id | item | count
---------------------------
1 | item 1 | 2
3 | item 4 | 2
4 | item 4 | 4
5 | item 5 | 2
I need to combine the two dataframes such that the existing items count are incremented and new items are inserted.
The result should be like:
Item Id | item | count
---------------------------
1 | item 1 | 4
2 | item 2 | 3
1 | item 3 | 2
3 | item 4 | 7
4 | item 4 | 4
5 | item 5 | 2
I have one way do achieve this not sure if its efficient or the right way to do
temp1 = df1.join(temp,['item_id','item'],'full_outer') \
.na.fill(0)
temp1\
.groupby("item_id", "item")\
.agg(F.sum(temp1["count"] + temp1["newcount"]))\
.show()

Since, the schema for the two dataframes is the same you can perform a union and then do a groupby id and aggregate the counts.
step1: df3 = df1.union(df2);
step2: df3.groupBy("Item Id", "item").agg(sum("count").as("count"));

There are several ways how to do it.
Based on what you describe the most straightforward solution would be to use RDD - SparkContext.union:
rdd1 = sc.parallelize(DF1)
rdd2 = sc.parallelize(DF2)
union_rdd = sc.union([rdd1, rdd2])
the alternative solution would be to use DataFrame.union from pyspark.sql
Note: I have suggested unionAll previously but it is deprecated in Spark 2.0

#wandermonk's solution is recommended as it does not use join. Avoid joins as much as possible as this triggers shuffling (also known as wide transformation and leads to data transfer over the network and that is expensive and slow)
You also have to look into your data size (both tables are big or one small one big etc) and accordingly you can tune the performance side of it.
I tried showing the group by a solution using SparkSQL as they do the same thing but easier to understand and manipulate.
from pyspark.sql.types import StructType, StructField, IntegerType, StringType
list_1 = [[1,"item 1" , 2],[2 ,"item 2", 3],[1 ,"item 3" ,2],[3 ,"item 4" , 5]]
list_2 = [[1,"item 1",2],[3 ,"item 4",2],[4 ,"item 4",4],[5 ,"item 5",2]]
my_schema = StructType([StructField("Item_ID",IntegerType(), True),StructField("Item_Name",StringType(), True ),StructField("Quantity",IntegerType(), True)])
df1 = spark.createDataFrame(list_1, my_schema)
df2 = spark.createDataFrame(list_2, my_schema)
df1.createOrReplaceTempView("df1")
df1.createOrReplaceTempView("df2")
df3 = df2.union(df1)
df3.createOrReplaceTempView("df3")
df4 = spark.sql("select Item_ID, Item_Name, sum(Quantity) as Quantity from df3 group by Item_ID, Item_Name")
df4.show(10)
now if you look into the SparkUI, you can see for such a small data set, the shuffle operation, and # of stages.
Number of stages for such a small job
Number the shuffle operation for this group by command
I also recommend to see the SQL plan and understand the cost. Exchange represents the shuffle here.
== Physical Plan ==
*(2) HashAggregate(keys=[Item_ID#6, Item_Name#7], functions=[sum(cast(Quantity#8 as bigint))], output=[Item_ID#6, Item_Name#7, Quantity#32L])
+- Exchange hashpartitioning(Item_ID#6, Item_Name#7, 200)
+- *(1) HashAggregate(keys=[Item_ID#6, Item_Name#7], functions=[partial_sum(cast(Quantity#8 as bigint))], output=[Item_ID#6, Item_Name#7, sum#38L])
+- Union
:- Scan ExistingRDD[Item_ID#6,Item_Name#7,Quantity#8]
+- Scan ExistingRDD[Item_ID#0,Item_Name#1,Quantity#2]