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Pyspark: Reorder only a subset of rows among themselves

my data frame:
+-----+--------+-------+
| val | id | reRnk |
+-----+--------+-------+
| 2 | a | yes |
| 1 | b | no |
| 3 | c | no |
| 8 | d | yes |
| 7 | e | yes |
| 9 | f | no |
+-----+--------+-------+
In my desired output I will re-rank only the columns where reRnk==yes, ranking will be done based on "val"
I don't want to change the rows where reRnk = no, for example at id=b we have reRnk=no I want to keep that row at row no. 2 only.
my desired output will look like this:
+-----+--------+-------+
| val | id | reRnk |
+-----+--------+-------+
| 8 | d | yes |
| 1 | b | no |
| 3 | c | no |
| 7 | e | yes |
| 2 | a | yes |
| 9 | f | no |
+-----+--------+-------+

From what I'm reading, pyspark DF's do not have an index by default. You might need to add this.
I do not know the exact syntax for pyspark, however since it has many similarities with pandas this might lead you into a certain direction:
df.loc[df.reRnk == 'yes', ['val','id']] = df.loc[df.reRnk == 'yes', ['val','id']].sort_values('val', ascending=False).set_index(df.loc[df.reRnk == 'yes', ['val','id']].index)
Basically what we do here is isolating the rows with reRnk == 'yes', sorting these values but resetting the index to its original index. Then we assign these new values to the original rows in the df.
for .loc, https://spark.apache.org/docs/3.2.0/api/python/reference/pyspark.pandas/api/pyspark.pandas.DataFrame.loc.html might be worth a try.
for .sort_values see: https://sparkbyexamples.com/pyspark/pyspark-orderby-and-sort-explained/

Pandas Coalesce Multiple Columns, NaN

I want to coalesce 4 columns using pandas. I've tried this:
final['join_key'] = final['book'].astype('str') + final['bdr'] + final['cusip'].fillna(final['isin']).fillna(final['Deal'].astype('str')).fillna(final['Id'])
When I use this it returns:
+-------+--------+-------+------+------+------------+------------------+
| book | bdr | cusip | isin | Deal | Id | join_key |
+-------+--------+-------+------+------+------------+------------------+
| 17236 | ETFROS | | | | 8012398421 | 17236.0ETFROSnan |
+-------+--------+-------+------+------+------------+------------------+
The field Id is not properly appending to my join_key field.
Any help would be appreciated, thanks.
Update:
+------------+------+------+-----------+--------------+------+------------+----------------------------+
| endOfDay | book | bdr | cusip | isin | Deal | Id | join_key |
+------------+------+------+-----------+--------------+------+------------+----------------------------+
| 31/10/2019 | 15 | ITOR | 371494AM7 | US371494AM77 | 161 | 8013210731 | 20191031|15|ITOR|371494AM7 |
| 31/10/2019 | 15 | ITOR | | | | 8011898573 | 20191031|15|ITOR| |
| 31/10/2019 | 15 | ITOR | | | | 8011898742 | 20191031|15|ITOR| |
| 31/10/2019 | 15 | ITOR | | | | 8011899418 | 20191031|15|ITOR| |
+------------+------+------+-----------+--------------+------+------------+----------------------------+
df['join_key'] = ("20191031|" + df['book'].astype('str') + "|" + df['bdr'] + "|" + df[['cusip', 'isin', 'Deal', 'id']].bfill(1)['cusip'].astype(str))
For some reason this code isnt picking up Id as part of the key.

The last chain fillna for cusip is too complicated. You may change it to bfill
final['join_key'] = (final['book'].astype('str') +
final['bdr'] +
final[['cusip', 'isin', 'Deal', 'Id']].bfill(1)['cusip'].astype(str))

Try this:
import pandas as pd
import numpy as np
# setup (ignore)
final = pd.DataFrame({
'book': [17236],
'bdr': ['ETFROS'],
'cusip': [np.nan],
'isin': [np.nan],
'Deal': [np.nan],
'Id': ['8012398421'],
})
# answer
final['join_key'] = final['book'].astype('str') + final['bdr'] + final['cusip'].fillna(final['isin']).fillna(final['Deal']).fillna(final['Id']).astype('str')
Output
book bdr cusip isin Deal Id join_key
0 17236 ETFROS NaN NaN NaN 8012398421 17236ETFROS8012398421

How to aggregate and restructure dataframe data in pyspark (column wise)

I am trying to aggregate data in pyspark dataframe on a particular criteria. I am trying to align the acct based on switchOUT amount to switchIN amount. So that accounts with money switching out of becomes from account and other accounts become to_accounts.
Data I am getting in the dataframe to begin with
+--------+------+-----------+----------+----------+-----------+
| person | acct | close_amt | open_amt | switchIN | switchOUT |
+--------+------+-----------+----------+----------+-----------+
| A | 1 | 125 | 50 | 75 | 0 |
+--------+------+-----------+----------+----------+-----------+
| A | 2 | 100 | 75 | 25 | 0 |
+--------+------+-----------+----------+----------+-----------+
| A | 3 | 200 | 300 | 0 | 100 |
+--------+------+-----------+----------+----------+-----------+
To this table
+--------+--------+-----------+----------+----------+
| person | from_acct| to_acct | switchIN | switchOUT|
+--------+----------+--------+----------+-----------+
| A | 3 | 1 | 75 | 100 |
+--------+----------+--------+----------+-----------+
| A | 3 | 2 | 25 | 100 |
+--------+----------+--------+----------+-----------+
And also how can I do it so that it works for N number of rows (not just 3 accounts)
So far I have used this code
# define udf
def sorter(l):
res = sorted(l, key=operator.itemgetter(1))
return [item[0] for item in res]
def list_to_string(l):
res = 'from_fund_' +str(l[0]) + '_to_fund_'+str(l[1])
return res
def listfirstAcc(l):
res = str(l[0])
return res
def listSecAcc(l):
res = str(l[1])
return res
sort_udf = F.udf(sorter)
list_str = F.udf(list_to_string)
extractFirstFund = F.udf(listfirstAcc)
extractSecondFund = F.udf(listSecAcc)
# Add additional columns
df= df.withColumn("move", sort_udf("list_col").alias("sorted_list"))
df= df.withColumn("move_string", list_str("move"))
df= df.withColumn("From_Acct",extractFirstFund("move"))
df= df.withColumn("To_Acct",extractSecondFund("move"))
Current outcome I am getting:
+--------+--------+-----------+----------+----------+
| person | from_acct| to_acct | switchIN | switchOUT|
+--------+----------+--------+----------+-----------+
| A | 3 | 1,2 | 75 | 100 |
+--------+----------+--------+----------+-----------+

Python: decrease running time of for loops

I want to calculate APRU for several countries.
country_list = ['us','gb','ca','id']
count = {}
for i in country_list:
count[i] = df_day_country[df_day_country.isin([i])]
count[i+'_reverse'] = count[i].iloc[::-1]
for j in range(1,len(count[i+'_reverse'])):
count[i+'_reverse']['count'].iloc[j] = count[i+'_reverse']['count'][j-1:j+1].sum()
for k in range(1,len(count[i])):
count[i][revenue_sum].iloc[k] = count[i][revenue_sum][k-1:k+1].sum()
count[i]['APRU'] = count[i][revenue_sum] / count[i]['count'][0]/100
After that, I will create 4 dataframes: df_us, df_gb, df_ca, df_id that show each country's APRU.
But the size of dataset is large. The running time is extremely slow after the country list become larger. So is there a way to decrease the running time?

Consider using numba
Your code thus becomes
from numba import njit
country_list = ['us','gb','ca','id']
#njit
def count(country_list):
count = {}
for i in country_list:
count[i] = df_day_country[df_day_country.isin([i])]
count[i+'_reverse'] = count[i].iloc[::-1]
for j in range(1,len(count[i+'_reverse'])):
count[i+'_reverse']['count'].iloc[j] = count[i+'_reverse']['count'][j-1:j+1].sum()
for k in range(1,len(count[i])):
count[i][revenue_sum].iloc[k] = count[i][revenue_sum][k-1:k+1].sum()
count[i]['APRU'] = count[i][revenue_sum] / count[i]['count'][0]/100
return count
Numba makes python loops a lot faster and is in the process of being integrated into the more heavy duty python libraries like scipy. Deffinetly give this a look.

IIUC, from your code and variable names, it looks like you are trying to compute average:
# toy data set:
country_list = ['us','gb']
np.random.seed(1)
datalen=10
df_day_country = pd.DataFrame({'country': np.random.choice(country_list, datalen),
'count': np.random.randint(0,100, datalen),
'revenue_sum': np.random.uniform(0,100,datalen)})
df_day_country['APRU'] = (df_day_country.groupby('country',group_keys=False)
.apply(lambda x: x['revenue_sum']/x['count'].sum())
)
Output:
+----------+--------+--------------+------------+----------+
| country | count | revenue_sum | APRU | |
+----------+--------+--------------+------------+----------+
| 0 | gb | 16 | 20.445225 | 0.150333 |
| 1 | gb | 1 | 87.811744 | 0.645675 |
| 2 | us | 76 | 2.738759 | 0.011856 |
| 3 | us | 71 | 67.046751 | 0.290246 |
| 4 | gb | 6 | 41.730480 | 0.306842 |
| 5 | gb | 25 | 55.868983 | 0.410801 |
| 6 | gb | 50 | 14.038694 | 0.103226 |
| 7 | gb | 20 | 19.810149 | 0.145663 |
| 8 | gb | 18 | 80.074457 | 0.588783 |
| 9 | us | 84 | 96.826158 | 0.419161 |
+----------+--------+--------------+------------+----------+

Convert graphlab sframe into a dictionary of {key: values}

Given an SFrame as such:
+------+-----------+-----------+-----------+-----------+-----------+-----------+
| X1 | X2 | X3 | X4 | X5 | X6 | X7 |
+------+-----------+-----------+-----------+-----------+-----------+-----------+
| the | -0.060292 | 0.06763 | -0.036891 | 0.066684 | 0.024045 | 0.099091 |
| , | 0.026625 | 0.073101 | -0.027073 | -0.019504 | 0.04173 | 0.038811 |
| . | -0.005893 | 0.093791 | 0.015333 | 0.046226 | 0.032791 | 0.110069 |
| of | -0.050371 | 0.031452 | 0.04091 | 0.033255 | -0.009195 | 0.061086 |
| and | 0.005456 | 0.063237 | -0.075793 | -0.000819 | 0.003407 | 0.053554 |
| to | 0.01347 | 0.043712 | -0.087122 | 0.015258 | 0.08834 | 0.139644 |
| in | -0.019466 | 0.077509 | -0.102543 | 0.034337 | 0.130886 | 0.032195 |
| a | -0.072288 | -0.017494 | -0.018383 | 0.001857 | -0.04645 | 0.133424 |
| is | 0.052726 | 0.041903 | 0.163781 | 0.006887 | -0.07533 | 0.108394 |
| for | -0.004082 | -0.024244 | 0.042166 | 0.007032 | -0.081243 | 0.026162 |
| on | -0.023709 | -0.038306 | -0.16072 | -0.171599 | 0.150983 | 0.042044 |
| that | 0.062037 | 0.100348 | -0.059753 | -0.041444 | 0.041156 | 0.166704 |
| ) | 0.052312 | 0.072473 | -0.02067 | -0.015581 | 0.063368 | -0.017216 |
| ( | 0.051408 | 0.186162 | 0.03028 | -0.048425 | 0.051376 | 0.004989 |
| with | 0.091825 | -0.081649 | -0.087926 | -0.061273 | 0.043528 | 0.107864 |
| was | 0.046042 | -0.058529 | 0.040581 | 0.067748 | 0.053724 | 0.041067 |
| as | 0.025248 | -0.012519 | -0.054685 | -0.040581 | 0.051061 | 0.114956 |
| it | 0.028606 | 0.106391 | 0.025065 | 0.023486 | 0.011184 | 0.016715 |
| by | -0.096704 | 0.150165 | -0.01775 | -0.07178 | 0.004458 | 0.098807 |
| be | -0.109489 | -0.025908 | 0.025608 | 0.076263 | -0.047246 | 0.100489 |
+------+-----------+-----------+-----------+-----------+-----------+-----------+
How can I convert the SFrame into a dictionary such that X1 column is the key and X2 to X7 as the np.array()?
I have tried iterating through the original SFrame row-by-row and do something like this:
>>> import graphlab as gl
>>> import numpy as np
>>> x = gl.SFrame()
>>> a = np.array([1,2,3])
>>> w = 'foo'
>>> x.append(gl.SFrame({'word':[w], 'vector':[a]}))
Columns:
vector array
word str
Rows: 1
Data:
+-----------------+------+
| vector | word |
+-----------------+------+
| [1.0, 2.0, 3.0] | foo |
+-----------------+------+
[1 rows x 2 columns]
Is there another way to do the same?
EDITED
After trying #papayawarrior solution, it works if I can load the whole dataframe into memory but there's a few quriks that makes it odd.
Assuming that my original input to the SFrame is as presented above (with 501 columns) but in .csv file, I have the code to read them into the desired dictionary:
def get_embeddings(embedding_gzip, size):
coltypes = [str] + [float] * size
sf = gl.SFrame.read_csv('compose-vectors/' + embedding_gzip, delimiter='\t', column_type_hints=coltypes, header=False, quote_char='\0')
sf = sf.pack_columns(['X'+str(i) for i in range(2, size+1)])
df = sf.to_dataframe().set_index('X1')
print list(df)
return df.to_dict(orient='dict')['X2']
But oddly it gives this error:
File "sts_compose.py", line 28, in get_embeddings
return df.to_dict(orient='dict')['X2']
KeyError: 'X2'
So when I check for the column names before the conversion to dictionary, I found that my column names are not 'X1' and 'X2' but list(df) prints ['X501', 'X3'].
Is there something wrong with how I have converting the graphlab.SFrame -> pandas.DataFrame -> dict?
I know I can resolve the problem by doing this instead, but the question remains, "How did the column names become so strange?":
def get_embeddings(embedding_gzip, size):
coltypes = [str] + [float] * size
sf = gl.SFrame.read_csv('compose-vectors/' + embedding_gzip, delimiter='\t', column_type_hints=coltypes, header=False, quote_char='\0')
sf = sf.pack_columns(['X'+str(i) for i in range(2, size+1)])
df = sf.to_dataframe().set_index('X1')
col_names = list(df)
return df.to_dict(orient='dict')[col_names[1]]

Is there another way to do the same?
Yes, you can use the pack_columns method from the SFrame class.
import graphlab as gl
data = gl.SFrame()
data.add_column(gl.SArray(['foo', 'bar']), 'X1')
data.add_column(gl.SArray([1., 3.]), 'X2')
data.add_column(gl.SArray([2., 4.]), 'X3')
print data
+-----+-----+-----+
| X1 | X2 | X3 |
+-----+-----+-----+
| foo | 1.0 | 2.0 |
| bar | 3.0 | 4.0 |
+-----+-----+-----+
[2 rows x 3 columns]
import array
data = data.pack_columns(['X2', 'X3'], dtype=array.array, new_column_name='vector')
data = data.rename({'X1':'word'})
print data
+------+------------+
| word | vector |
+------+------------+
| foo | [1.0, 2.0] |
| bar | [3.0, 4.0] |
+------+------------+
[2 rows x 2 columns]
b=data['vector'][0]
print type(b)
<type 'array.array'>
How can I convert the SFrame into a dictionary such that X1 column is the key and X2 to X7 as the np.array()?
I didn't find any built-in method to convert an SFrame to a dict. You could try the following (it might be very slow):
a={}
def dump_sframe_to_dict(row, a):
a[row['word']]=row['vector']
data.apply(lambda x: dump_sframe_to_dict(x, a))
print a
{'foo': array('d', [1.0, 2.0]), 'bar': array('d', [3.0, 4.0])}

Edited to match new questions in the post.
#Adrien Renaud is spot on with the SFrame.pack_columns method, but I would suggest using the Pandas dataframe to_dict for the last question if your dataset fits in memory.
>>> import graphlab as gl
>>> sf = gl.SFrame({'X1': ['cat', 'dog'], 'X2': [1, 2], 'X3': [3, 4]})
>>> sf
+-----+----+----+
| X1 | X2 | X3 |
+-----+----+----+
| cat | 1 | 3 |
| dog | 2 | 4 |
+-----+----+----+
>>> sf2 = sf.rename({'X1': 'word'})
>>> sf2 = sf.pack_columns(column_prefix='X', new_column_name='vector')
>>> sf2
+------+--------+
| word | vector |
+------+--------+
| cat | [1, 3] |
| dog | [2, 4] |
+------+--------+
>>> df = sf2.to_dataframe().set_index('word')
>>> result = df.to_dict(orient='dict')['vector']
>>> result
{'cat': [1, 3], 'dog': [2, 4]}