Where does the leading whitespace come from in this RDD and how to avoid it? - apache-spark

string_integers.txt
a 1 2 3
b 4 5 6
c 7 8 9
sample.py
import re
pattern = re.compile("(^[a-z]+)\s")
txt = sc.textFile("string_integers.txt")
string_integers_separated = txt.map(lambda x: pattern.split(x))
print string_integers_separated.collect()
outcome
[[u'', u'a', u'1 2 3'], [u'', u'b', u'4 5 6'], [u'', u'c', u'7 8 9']]
expected outcome
[[u'a', u'1 2 3'], [u'b', u'4 5 6'], [u'c', u'7 8 9']]


You split on pattern anchored in the beginning of the string so prefix will always be the empty string. You can for example use match:
pattern = re.compile("([a-z]+)\s+(.*$)")
pattern.match("a 1 2 3").groups()
# ('a', '1 2 3')
or lookbehind:
pattern = re.compile("(?<=a)\s")
pattern.split("a 1 2 3", maxsplit=1)
# ['a', '1 2 3']
or just split:
"a 1 2 3".split(maxsplit=1)
# ['a', '1 2 3']

Related

Replace string column endwiths specific substrings under condition of another column with Pandas

Given a dataset as follows:
id company name value
0 1 Finl Corp. 7
1 2 Fund Tr Corp 6
2 3 Inc Invt Fd 5
3 4 Govt Fd Inc. 3
4 5 Trinity Inc 5
Or:
[{'id': 1, 'company name': 'Finl Corp.', 'value': 7},
{'id': 2, 'company name': 'Fund Tr Corp', 'value': 6},
{'id': 3, 'company name': 'Inc Invt Fd', 'value': 5},
{'id': 4, 'company name': 'Govt Fd Inc.', 'value': 3},
{'id': 5, 'company name': 'Trinity Inc', 'value': 5}]
I need to replace if company name column's contents endwiths ['Corp.', 'Corp', 'Inc.', 'Inc'], while at same time value is >= 5
The expected result will be:
id company name value
0 1 Finl 7
1 2 Fund Tr 6
2 3 Inc Invt Fd 5
3 4 Govt Fd Inc. 3
4 5 Trinity 5
How could I acheive that in Pandas and regex?
Trial code with error: TypeError: replace() missing 1 required positional argument: 'repl'
mask = (df1['value'] >= 5)
df1.loc[mask, 'company_name_concise']= df1.loc[mask, 'company name'].str.replace(r'\bCorp.|Corp|Inc.|Inc$', regex=True)

You can change values in regex by add \s* for spaces with $ for end of strings:
mask = (df1['value'] >= 5)
L = ['Corp.', 'Corp', 'Inc.', 'Inc']
pat = '|'.join(f'\s*{x}$' for x in L)
df1.loc[mask, 'company name']= df1.loc[mask,'company name'].str.replace(pat,'',regex=True)
print (df1)
id company name value
0 1 Finl 7
1 2 Fund Tr 6
2 3 Inc Invt Fd 5
3 4 Govt Fd Inc. 3
4 5 Trinity 5

str.replace takes two arguments, the pattern and the replacement:
mask = (df1['value'] >= 5)
df1.loc[mask, 'company_name_concise']= df1.loc[mask, 'company name'].str.replace(r'\b(?:Corp\.?|Inc\.?)$', '', regex=True)
Note that the regex pattern you want here is:
\b word boundary
(?:
Corp\.? match Corp or Corp.
| OR
Inc\.? match Inc or Inc.
)
$ at the end of the company name

Or for length of code you could just directly modify the whole column and assign by index:
df.loc[df['value'] > 5, 'company name'] = df['company name'].str.replace(r'\bCorp.|Corp|Inc.|Inc$', '')
>>> df
id company name value
0 1 Finl 7
1 2 Fund Tr 6
2 3 Inc Invt Fd 5
3 4 Govt Fd Inc. 3
4 5 Trinity Inc 5
>>>
Or a solution with np.where:
>>> df['company name'] = np.where(df['value'] > 5, df['company name'].str.replace(r'\bCorp.|Corp|Inc.|Inc$', ''), df['company name'])
>>> df
id company name value
0 1 Finl 7
1 2 Fund Tr 6
2 3 Inc Invt Fd 5
3 4 Govt Fd Inc. 3
4 5 Trinity Inc 5
>>>

Using Python3.x rename a column name of a dataframe with user provided new column name in console window

How to rename a column name of my DataFrame with the 'user provided new column name in the console screen'.

Say you have a dataframe like this
>>> df
A B C
0 1 9 1
1 4 7 2
2 3 7 6
3 6 1 6
4 8 1 9
And you want to rename the column B to that given by user:
>>> col_name = input("Enter Column Name: ")
Enter Column Name: test
>>> col_names = df.columns.tolist()
>>> col_names
['A', 'B', 'C']
>>> col_names[1] = col_name
>>> col_names
['A', 'test', 'C']
>>> df.columns = col_names
>>> df
A test C
0 1 9 1
1 4 7 2
2 3 7 6
3 6 1 6
4 8 1 9

Select row by max of a column Pandas Python [duplicate]

How can I perform aggregation with Pandas?
No DataFrame after aggregation! What happened?
How can I aggregate mainly strings columns (to lists, tuples, strings with separator)?
How can I aggregate counts?
How can I create a new column filled by aggregated values?
I've seen these recurring questions asking about various faces of the pandas aggregate functionality.
Most of the information regarding aggregation and its various use cases today is fragmented across dozens of badly worded, unsearchable posts.
The aim here is to collate some of the more important points for posterity.
This Q&A is meant to be the next instalment in a series of helpful user-guides:
How to pivot a dataframe,
Pandas concat
How do I operate on a DataFrame with a Series for every column?
Pandas Merging 101
Please note that this post is not meant to be a replacement for the documentation about aggregation and about groupby, so please read that as well!

Question 1
How can I perform aggregation with Pandas?
Expanded aggregation documentation.
Aggregating functions are the ones that reduce the dimension of the returned objects. It means output Series/DataFrame have less or same rows like original.
Some common aggregating functions are tabulated below:
Function Description
mean() Compute mean of groups
sum() Compute sum of group values
size() Compute group sizes
count() Compute count of group
std() Standard deviation of groups
var() Compute variance of groups
sem() Standard error of the mean of groups
describe() Generates descriptive statistics
first() Compute first of group values
last() Compute last of group values
nth() Take nth value, or a subset if n is a list
min() Compute min of group values
max() Compute max of group values
np.random.seed(123)
df = pd.DataFrame({'A' : ['foo', 'foo', 'bar', 'foo', 'bar', 'foo'],
'B' : ['one', 'two', 'three','two', 'two', 'one'],
'C' : np.random.randint(5, size=6),
'D' : np.random.randint(5, size=6),
'E' : np.random.randint(5, size=6)})
print (df)
A B C D E
0 foo one 2 3 0
1 foo two 4 1 0
2 bar three 2 1 1
3 foo two 1 0 3
4 bar two 3 1 4
5 foo one 2 1 0
Aggregation by filtered columns and Cython implemented functions:
df1 = df.groupby(['A', 'B'], as_index=False)['C'].sum()
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
An aggregate function is used for all columns without being specified in the groupby function, here the A, B columns:
df2 = df.groupby(['A', 'B'], as_index=False).sum()
print (df2)
A B C D E
0 bar three 2 1 1
1 bar two 3 1 4
2 foo one 4 4 0
3 foo two 5 1 3
You can also specify only some columns used for aggregation in a list after the groupby function:
df3 = df.groupby(['A', 'B'], as_index=False)['C','D'].sum()
print (df3)
A B C D
0 bar three 2 1
1 bar two 3 1
2 foo one 4 4
3 foo two 5 1
Same results by using function DataFrameGroupBy.agg:
df1 = df.groupby(['A', 'B'], as_index=False)['C'].agg('sum')
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
df2 = df.groupby(['A', 'B'], as_index=False).agg('sum')
print (df2)
A B C D E
0 bar three 2 1 1
1 bar two 3 1 4
2 foo one 4 4 0
3 foo two 5 1 3
For multiple functions applied for one column use a list of tuples - names of new columns and aggregated functions:
df4 = (df.groupby(['A', 'B'])['C']
.agg([('average','mean'),('total','sum')])
.reset_index())
print (df4)
A B average total
0 bar three 2.0 2
1 bar two 3.0 3
2 foo one 2.0 4
3 foo two 2.5 5
If want to pass multiple functions is possible pass list of tuples:
df5 = (df.groupby(['A', 'B'])
.agg([('average','mean'),('total','sum')]))
print (df5)
C D E
average total average total average total
A B
bar three 2.0 2 1.0 1 1.0 1
two 3.0 3 1.0 1 4.0 4
foo one 2.0 4 2.0 4 0.0 0
two 2.5 5 0.5 1 1.5 3
Then get MultiIndex in columns:
print (df5.columns)
MultiIndex(levels=[['C', 'D', 'E'], ['average', 'total']],
labels=[[0, 0, 1, 1, 2, 2], [0, 1, 0, 1, 0, 1]])
And for converting to columns, flattening MultiIndex use map with join:
df5.columns = df5.columns.map('_'.join)
df5 = df5.reset_index()
print (df5)
A B C_average C_total D_average D_total E_average E_total
0 bar three 2.0 2 1.0 1 1.0 1
1 bar two 3.0 3 1.0 1 4.0 4
2 foo one 2.0 4 2.0 4 0.0 0
3 foo two 2.5 5 0.5 1 1.5 3
Another solution is pass list of aggregate functions, then flatten MultiIndex and for another columns names use str.replace:
df5 = df.groupby(['A', 'B']).agg(['mean','sum'])
df5.columns = (df5.columns.map('_'.join)
.str.replace('sum','total')
.str.replace('mean','average'))
df5 = df5.reset_index()
print (df5)
A B C_average C_total D_average D_total E_average E_total
0 bar three 2.0 2 1.0 1 1.0 1
1 bar two 3.0 3 1.0 1 4.0 4
2 foo one 2.0 4 2.0 4 0.0 0
3 foo two 2.5 5 0.5 1 1.5 3
If want specified each column with aggregated function separately pass dictionary:
df6 = (df.groupby(['A', 'B'], as_index=False)
.agg({'C':'sum','D':'mean'})
.rename(columns={'C':'C_total', 'D':'D_average'}))
print (df6)
A B C_total D_average
0 bar three 2 1.0
1 bar two 3 1.0
2 foo one 4 2.0
3 foo two 5 0.5
You can pass custom function too:
def func(x):
return x.iat[0] + x.iat[-1]
df7 = (df.groupby(['A', 'B'], as_index=False)
.agg({'C':'sum','D': func})
.rename(columns={'C':'C_total', 'D':'D_sum_first_and_last'}))
print (df7)
A B C_total D_sum_first_and_last
0 bar three 2 2
1 bar two 3 2
2 foo one 4 4
3 foo two 5 1
Question 2
No DataFrame after aggregation! What happened?
Aggregation by two or more columns:
df1 = df.groupby(['A', 'B'])['C'].sum()
print (df1)
A B
bar three 2
two 3
foo one 4
two 5
Name: C, dtype: int32
First check the Index and type of a Pandas object:
print (df1.index)
MultiIndex(levels=[['bar', 'foo'], ['one', 'three', 'two']],
labels=[[0, 0, 1, 1], [1, 2, 0, 2]],
names=['A', 'B'])
print (type(df1))
<class 'pandas.core.series.Series'>
There are two solutions for how to get MultiIndex Series to columns:
add parameter as_index=False
df1 = df.groupby(['A', 'B'], as_index=False)['C'].sum()
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
use Series.reset_index:
df1 = df.groupby(['A', 'B'])['C'].sum().reset_index()
print (df1)
A B C
0 bar three 2
1 bar two 3
2 foo one 4
3 foo two 5
If group by one column:
df2 = df.groupby('A')['C'].sum()
print (df2)
A
bar 5
foo 9
Name: C, dtype: int32
... get Series with Index:
print (df2.index)
Index(['bar', 'foo'], dtype='object', name='A')
print (type(df2))
<class 'pandas.core.series.Series'>
And the solution is the same like in the MultiIndex Series:
df2 = df.groupby('A', as_index=False)['C'].sum()
print (df2)
A C
0 bar 5
1 foo 9
df2 = df.groupby('A')['C'].sum().reset_index()
print (df2)
A C
0 bar 5
1 foo 9
Question 3
How can I aggregate mainly strings columns (to lists, tuples, strings with separator)?
df = pd.DataFrame({'A' : ['a', 'c', 'b', 'b', 'a', 'c', 'b'],
'B' : ['one', 'two', 'three','two', 'two', 'one', 'three'],
'C' : ['three', 'one', 'two', 'two', 'three','two', 'one'],
'D' : [1,2,3,2,3,1,2]})
print (df)
A B C D
0 a one three 1
1 c two one 2
2 b three two 3
3 b two two 2
4 a two three 3
5 c one two 1
6 b three one 2
Instead of an aggregation function, it is possible to pass list, tuple, set for converting the column:
df1 = df.groupby('A')['B'].agg(list).reset_index()
print (df1)
A B
0 a [one, two]
1 b [three, two, three]
2 c [two, one]
An alternative is use GroupBy.apply:
df1 = df.groupby('A')['B'].apply(list).reset_index()
print (df1)
A B
0 a [one, two]
1 b [three, two, three]
2 c [two, one]
For converting to strings with a separator, use .join only if it is a string column:
df2 = df.groupby('A')['B'].agg(','.join).reset_index()
print (df2)
A B
0 a one,two
1 b three,two,three
2 c two,one
If it is a numeric column, use a lambda function with astype for converting to strings:
df3 = (df.groupby('A')['D']
.agg(lambda x: ','.join(x.astype(str)))
.reset_index())
print (df3)
A D
0 a 1,3
1 b 3,2,2
2 c 2,1
Another solution is converting to strings before groupby:
df3 = (df.assign(D = df['D'].astype(str))
.groupby('A')['D']
.agg(','.join).reset_index())
print (df3)
A D
0 a 1,3
1 b 3,2,2
2 c 2,1
For converting all columns, don't pass a list of column(s) after groupby.
There isn't any column D, because automatic exclusion of 'nuisance' columns. It means all numeric columns are excluded.
df4 = df.groupby('A').agg(','.join).reset_index()
print (df4)
A B C
0 a one,two three,three
1 b three,two,three two,two,one
2 c two,one one,two
So it's necessary to convert all columns into strings, and then get all columns:
df5 = (df.groupby('A')
.agg(lambda x: ','.join(x.astype(str)))
.reset_index())
print (df5)
A B C D
0 a one,two three,three 1,3
1 b three,two,three two,two,one 3,2,2
2 c two,one one,two 2,1
Question 4
How can I aggregate counts?
df = pd.DataFrame({'A' : ['a', 'c', 'b', 'b', 'a', 'c', 'b'],
'B' : ['one', 'two', 'three','two', 'two', 'one', 'three'],
'C' : ['three', np.nan, np.nan, 'two', 'three','two', 'one'],
'D' : [np.nan,2,3,2,3,np.nan,2]})
print (df)
A B C D
0 a one three NaN
1 c two NaN 2.0
2 b three NaN 3.0
3 b two two 2.0
4 a two three 3.0
5 c one two NaN
6 b three one 2.0
Function GroupBy.size for size of each group:
df1 = df.groupby('A').size().reset_index(name='COUNT')
print (df1)
A COUNT
0 a 2
1 b 3
2 c 2
Function GroupBy.count excludes missing values:
df2 = df.groupby('A')['C'].count().reset_index(name='COUNT')
print (df2)
A COUNT
0 a 2
1 b 2
2 c 1
This function should be used for multiple columns for counting non-missing values:
df3 = df.groupby('A').count().add_suffix('_COUNT').reset_index()
print (df3)
A B_COUNT C_COUNT D_COUNT
0 a 2 2 1
1 b 3 2 3
2 c 2 1 1
A related function is Series.value_counts. It returns the size of the object containing counts of unique values in descending order, so that the first element is the most frequently-occurring element. It excludes NaNs values by default.
df4 = (df['A'].value_counts()
.rename_axis('A')
.reset_index(name='COUNT'))
print (df4)
A COUNT
0 b 3
1 a 2
2 c 2
If you want same output like using function groupby + size, add Series.sort_index:
df5 = (df['A'].value_counts()
.sort_index()
.rename_axis('A')
.reset_index(name='COUNT'))
print (df5)
A COUNT
0 a 2
1 b 3
2 c 2
Question 5
How can I create a new column filled by aggregated values?
Method GroupBy.transform returns an object that is indexed the same (same size) as the one being grouped.
See the Pandas documentation for more information.
np.random.seed(123)
df = pd.DataFrame({'A' : ['foo', 'foo', 'bar', 'foo', 'bar', 'foo'],
'B' : ['one', 'two', 'three','two', 'two', 'one'],
'C' : np.random.randint(5, size=6),
'D' : np.random.randint(5, size=6)})
print (df)
A B C D
0 foo one 2 3
1 foo two 4 1
2 bar three 2 1
3 foo two 1 0
4 bar two 3 1
5 foo one 2 1
df['C1'] = df.groupby('A')['C'].transform('sum')
df['C2'] = df.groupby(['A','B'])['C'].transform('sum')
df[['C3','D3']] = df.groupby('A')['C','D'].transform('sum')
df[['C4','D4']] = df.groupby(['A','B'])['C','D'].transform('sum')
print (df)
A B C D C1 C2 C3 D3 C4 D4
0 foo one 2 3 9 4 9 5 4 4
1 foo two 4 1 9 5 9 5 5 1
2 bar three 2 1 5 2 5 2 2 1
3 foo two 1 0 9 5 9 5 5 1
4 bar two 3 1 5 3 5 2 3 1
5 foo one 2 1 9 4 9 5 4 4

If you are coming from an R or SQL background, here are three examples that will teach you everything you need to do aggregation the way you are already familiar with:
Let us first create a Pandas dataframe
import pandas as pd
df = pd.DataFrame({'key1' : ['a','a','a','b','a'],
'key2' : ['c','c','d','d','e'],
'value1' : [1,2,2,3,3],
'value2' : [9,8,7,6,5]})
df.head(5)
Here is how the table we created looks like:
key1
key2
value1
value2
a
c
1
9
a
c
2
8
a
d
2
7
b
d
3
6
a
e
3
5
1. Aggregating With Row Reduction Similar to SQL Group By
1.1 If Pandas version >=0.25
Check your Pandas version by running print(pd.__version__). If your Pandas version is 0.25 or above then the following code will work:
df_agg = df.groupby(['key1','key2']).agg(mean_of_value_1=('value1', 'mean'),
sum_of_value_2=('value2', 'sum'),
count_of_value1=('value1','size')
).reset_index()
df_agg.head(5)
The resulting data table will look like this:
key1
key2
mean_of_value1
sum_of_value2
count_of_value1
a
c
1.5
17
2
a
d
2.0
7
1
a
e
3.0
5
1
b
d
3.0
6
1
The SQL equivalent of this is:
SELECT
key1
,key2
,AVG(value1) AS mean_of_value_1
,SUM(value2) AS sum_of_value_2
,COUNT(*) AS count_of_value1
FROM
df
GROUP BY
key1
,key2
1.2 If Pandas version <0.25
If your Pandas version is older than 0.25 then running the above code will give you the following error:
TypeError: aggregate() missing 1 required positional argument: 'arg'
Now to do the aggregation for both value1 and value2, you will run this code:
df_agg = df.groupby(['key1','key2'],as_index=False).agg({'value1':['mean','count'],'value2':'sum'})
df_agg.columns = ['_'.join(col).strip() for col in df_agg.columns.values]
df_agg.head(5)
The resulting table will look like this:
key1
key2
value1_mean
value1_count
value2_sum
a
c
1.5
2
17
a
d
2.0
1
7
a
e
3.0
1
5
b
d
3.0
1
6
Renaming the columns needs to be done separately using the below code:
df_agg.rename(columns={"value1_mean" : "mean_of_value1",
"value1_count" : "count_of_value1",
"value2_sum" : "sum_of_value2"
}, inplace=True)
2. Create a Column Without Reduction in Rows (EXCEL - SUMIF, COUNTIF)
If you want to do a SUMIF, COUNTIF, etc., like how you would do in Excel where there is no reduction in rows, then you need to do this instead.
df['Total_of_value1_by_key1'] = df.groupby('key1')['value1'].transform('sum')
df.head(5)
The resulting data frame will look like this with the same number of rows as the original:
key1
key2
value1
value2
Total_of_value1_by_key1
a
c
1
9
8
a
c
2
8
8
a
d
2
7
8
b
d
3
6
3
a
e
3
5
8
3. Creating a RANK Column ROW_NUMBER() OVER (PARTITION BY ORDER BY)
Finally, there might be cases where you want to create a rank column which is the SQL equivalent of ROW_NUMBER() OVER (PARTITION BY key1 ORDER BY value1 DESC, value2 ASC).
Here is how you do that.
df['RN'] = df.sort_values(['value1','value2'], ascending=[False,True]) \
.groupby(['key1']) \
.cumcount() + 1
df.head(5)
Note: we make the code multi-line by adding \ at the end of each line.
Here is how the resulting data frame looks like:
key1
key2
value1
value2
RN
a
c
1
9
4
a
c
2
8
3
a
d
2
7
2
b
d
3
6
1
a
e
3
5
1
In all the examples above, the final data table will have a table structure and won't have the pivot structure that you might get in other syntaxes.
Other aggregating operators:
mean() Compute mean of groups
sum() Compute sum of group values
size() Compute group sizes
count() Compute count of group
std() Standard deviation of groups
var() Compute variance of groups
sem() Standard error of the mean of groups
describe() Generates descriptive statistics
first() Compute first of group values
last() Compute last of group values
nth() Take nth value, or a subset if n is a list
min() Compute min of group values
max() Compute max of group values

Pandas aggregate column and keep header

I have code which works but gives me data without header is there a way I can write this code so header is not removed? I know one way will be to add back header, but is there a better way?
My code:
df = pd.read_csv(“_data.csv",skiprows=[0], header=None)
df = df.groupby([2])[10].sum().astype(float)
Data:
A B
1 2
1 1
2 3
2 4
I have data like above trying to get this result:
A B
1 3
2 7

Try to use the function reset_index after the sum:
data = [{'a': 1, 'b': 2},{'a': 1, 'b': 1},{'a': 2, 'b': 3},{'a': 2, 'b': 4}]
df = pd.DataFrame(data)
df
a b
0 1 2
1 1 1
2 2 3
3 2 4
df.groupby('a').sum().reset_index()
a b
0 1 3
1 2 7

You should specify the separator (several spaces in your case) and that the header is the first row (=0, with python indexing), than groupby the column you want.
df = pd.read_csv("_data.csv", sep='\s*', header=0)
A B
0 1 2
1 1 1
2 2 3
3 2 4
df = df.groupby(['A']).sum()
B
A
1 3
2 7

Getting all rows where for column 'C' the entry is larger than the preceding element in column 'C'

How can I select all rows of a data frame where a condition is met according to a column, which has to do with the relationship between every 2 entries of that column. To give the specific example, lets say I have a DataFrame:
>>>df = pd.DataFrame({'A': [ 1, 2, 3, 4],
'B':['spam', 'ham', 'egg', 'foo'],
'C':[4, 5, 3, 4]})
>>> df
A B C
0 1 spam 4
1 2 ham 5
2 3 egg 3
3 4 foo 4
>>>df2 = df[ return every row of df where C[i] > C[i-1] ]
>>> df2
A B C
1 2 ham 5
3 4 foo 4
There is plenty of great information about slicing and indexing in the pandas docs and here, but this is a bit more complicated, I think. I could also be going about it wrong. What I'm looking for is the rows of data where the value stored in C is no longer monotonously declining.
Any help is appreciated!

Use boolean indexing with compare by shifted column values:
print (df[df['C'] > df['C'].shift()])
A B C
1 2 ham 5
3 4 foo 4
Detail:
print (df['C'] > df['C'].shift())
0 False
1 True
2 False
3 True
Name: C, dtype: bool
If want all monotonously declining rows compare diff of column:
print (df[df['C'].diff() > 0])
A B C
1 2 ham 5
3 4 foo 4

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