Develop Reference

Trailing empty string after re.split()

I have two strings where I want to isolate sequences of digits from everything else.
For example:
import re
s = 'abc123abc'
print(re.split('(\d+)', s))
s = 'abc123abc123'
print(re.split('(\d+)', s))
The output looks like this:
['abc', '123', 'abc']
['abc', '123', 'abc', '123', '']
Note that in the second case, there's a trailing empty string.
Obviously I can test for that and remove it if necessary but it seems cumbersome and I wondered if the RE can be improved to account for this scenario.

You can use filter and don't return this empty string like below:
>>> s = 'abc123abc123'
>>> re.split('(\d+)', s)
['abc', '123', 'abc', '123', '']
>>> list(filter(None,re.split('(\d+)', s)))
['abc', '123', 'abc', '123']
By thanks #chepner you can generate list comprehension like below:
>>> [x for x in re.split('(\d+)', s) if x]
['abc', '123', 'abc', '123']
If maybe you have symbols or other you need split:
>>> s = '&^%123abc123$##123'
>>> list(filter(None,re.split('(\d+)', s)))
['&^%', '123', 'abc', '123', '$##', '123']

This has to do with the implementation of re.split() itself: you can't change it. When the function splits, it doesn't check anything that comes after the capture group, so it can't choose for you to either keep or discard the empty string that is left after splitting. It just splits there and leaves the rest of the string (which can be empty) to the next cycle.
If you don't want that empty string, you can get rid of it in various ways before collecting the results into a list. user1740577's is one example, but personally I prefer a list comprehension, since it's more idiomatic for simple filter/map operations:
parts = [part for part in re.split('(\d+)', s) if part]
I recommend against checking and getting rid of the element after the list has already been created, because it involves more operations and allocations.

A simple way to use regular expressions for this would be re.findall:
def bits(s):
return re.findall(r"(\D+|\d+)", s)
bits("abc123abc123")
# ['abc', '123', 'abc', '123']
But it seems easier and more natural with itertools.groupby. After all, you are chunking an iterable based on a single condition:
from itertools import groupby
def bits(s):
return ["".join(g) for _, g in groupby(s, key=str.isdigit)]
bits("abc123abc123")
# ['abc', '123', 'abc', '123']

Deleting data from dictionary based on starting letter

I have a data set with 4 columns, I have already opened, read, and made each column into a key/dictionary, I am trying to filter out any data which begins with a certain letter, e.g. for key DA, any value in this key with a certain starting letter (e.g. E) will result in the row being deleted. How can I go about doing this?

You can use the startswith function to check if a string starts with a certain letter. So in your case, it can be something like the following:
list_dictionary = [
{'KeyYa': 'abc', 'KeyDa': 'def', 'KeyBa': 'ghi', 'KeySa': 'jkl'},
{'KeyYa': 'abc', 'KeyDa': 'Edef', 'KeyBa': 'ghi', 'KeySa': 'jkl'},
{'KeyYa': 'abc', 'KeyDa': 'Gdef', 'KeyBa': 'ghi', 'KeySa': 'jkl'},
{'KeyYa': 'abc', 'KeyDa': 'Edef', 'KeyBa': 'ghi', 'KeySa': 'jkl'}
]
filtered = []
for line_dict in list_dictionary:
if not line_dict['KeyDa'].startswith('E'):
filtered.append(line_dict)
print(filtered)
This prints:
[{'KeyDa': 'def', 'KeyYa': 'abc', 'KeyBa': 'ghi', 'KeySa': 'jkl'}, {'KeyDa': 'Gdef', 'KeyYa': 'abc', 'KeyBa': 'ghi', 'KeySa': 'jkl'}]
If you're comfortable with the filter function and lambda, you can also do this concisely like this:
filtered = list(filter(lambda line: not line['KeyDa'].startswith('E'), list_dictionary))

How do you rotate an array of strings in Python?

In Python, given the following array of strings,
[ 'abc',
'def',
'ghi',
'jkl'
]
how do you transform it so it becomes,
[ 'jgda',
'kheb',
'lifc'
]

Using zip and str.join
Ex:
a = ['abc', 'def', 'ghi', 'jkl']
for i in zip(*a):
print("".join(i)[::-1])
Output:
jgda
kheb
lifc
[::-1] to reverse the string.

You could use numpy
import numpy as np
x = ['abc',
'def',
'ghi',
'jkl'
]
a = np.rot90([list(row) for row in x], 3)
result = [''.join(row) for row in a]
output:
[
'jgda',
'kheb',
'lifc'
]

How can I remove a specific number from a string?

here is the code:
aList = ['0.01', 'xyz', 'J0.01', 'abc', 'xyz'];
aList.remove('0.01');
print("List : ", aList)
here is the output:
List :
['xyz', 'J0.01', 'abc', 'xyz']
How can I remove the 0.01 attached to 'J0.01'? I would like to keep the J. Thanks for your time! =)

Seems like you want
aList = ['0.01', 'xyz', 'J0.01', 'abc', 'xyz'];
>>> [z.replace('0.01', '') for z in aList]
['', 'xyz', 'J', 'abc', 'xyz']
If you want to remove also empty strings/whitespaces,
>>> [z.replace('0.01', '') for z in aList if z.replace('0.01', '').strip()]
['xyz', 'J', 'abc', 'xyz']

Using re module:
import re
aList = ['0.01', 'xyz', 'J0.01', 'abc', 'xyz'];
print([i for i in (re.sub(r'\d+\.?\d*$', '', i) for i in aList) if i])
Prints:
['xyz', 'J', 'abc', 'xyz']
EDIT:
The regexp substitution re.sub(r'\d+\.?\d*$', '', i) will substitute every digit followed by dot (optional) and followed by any number of digits for empty string. The $ signifies that the digit should be at the end of the string.
So. e.g. the following matches are valid: "0.01", "0.", "0". Explanation on external site here.

Something like that can works:
l = ['0.01', 'xyz', 'J0.01', 'abc', 'xyz']
string = '0.01'
result = []
for x in l :
if string in x:
substring = x.replace(string,'')
if substring != "":
result.append(substring)
else:
result.append(x)
print(result)
try it, regards.

How to split a string containing digits and characters

I want to split a long string (containing digits and characters in it without any space) in to different substrings in Python?
>>> s = "abc123cde4567"
after split will get
['abc', '123', 'cde', '4567']
Thank you!

>>> import re
>>> re.findall("[a-z]+|[0-9]+", "abc123cde4567")
['abc', '123', 'cde', '4567']

Something different from a regex:
from itertools import groupby
from string import digits
s = "abc123cde4567"
print [''.join(g) for k, g in groupby(s, digits.__contains__)]
# ['abc', '123', 'cde', '4567']