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isolate data from long string [duplicate]

Suppose I had a string
string1 = "498results should get"
Now I need to get only integer values from the string like 498. Here I don't want to use list slicing because the integer values may increase like these examples:
string2 = "49867results should get"
string3 = "497543results should get"
So I want to get only integer values out from the string exactly in the same order. I mean like 498,49867,497543 from string1,string2,string3 respectively.
Can anyone let me know how to do this in a one or two lines?

>>> import re
>>> string1 = "498results should get"
>>> int(re.search(r'\d+', string1).group())
498
If there are multiple integers in the string:
>>> map(int, re.findall(r'\d+', string1))
[498]

An answer taken from ChristopheD here: https://stackoverflow.com/a/2500023/1225603
r = "456results string789"
s = ''.join(x for x in r if x.isdigit())
print int(s)
456789

Here's your one-liner, without using any regular expressions, which can get expensive at times:
>>> ''.join(filter(str.isdigit, "1234GAgade5312djdl0"))
returns:
'123453120'

if you have multiple sets of numbers then this is another option
>>> import re
>>> print(re.findall('\d+', 'xyz123abc456def789'))
['123', '456', '789']
its no good for floating point number strings though.

Iterator version
>>> import re
>>> string1 = "498results should get"
>>> [int(x.group()) for x in re.finditer(r'\d+', string1)]
[498]

>>> import itertools
>>> int(''.join(itertools.takewhile(lambda s: s.isdigit(), string1)))

With python 3.6, these two lines return a list (may be empty)
>>[int(x) for x in re.findall('\d+', your_string)]
Similar to
>>list(map(int, re.findall('\d+', your_string))

this approach uses list comprehension, just pass the string as argument to the function and it will return a list of integers in that string.
def getIntegers(string):
numbers = [int(x) for x in string.split() if x.isnumeric()]
return numbers
Like this
print(getIntegers('this text contains some numbers like 3 5 and 7'))
Output
[3, 5, 7]

def function(string):
final = ''
for i in string:
try:
final += str(int(i))
except ValueError:
return int(final)
print(function("4983results should get"))

Another option is to remove the trailing the letters using rstrip and string.ascii_lowercase (to get the letters):
import string
out = [int(s.replace(' ','').rstrip(string.ascii_lowercase)) for s in strings]
Output:
[498, 49867, 497543]

integerstring=""
string1 = "498results should get"
for i in string1:
if i.isdigit()==True
integerstring=integerstring+i
print(integerstring)

Python list to string loop

If I want to disassemble a string into a list, do some manipulation with the original decimal values, and then assemble the string from the list, what is the best way?
str = 'abc'
lst = list(str.encode('utf-8'))
for i in lst:
print (i, chr(int(i+2)))
gives me a table.
But I would like to create instead a presentation like 'abc', 'cde', etc.

Hope this helps
str_ini = 'abc'
lst = list(str_ini.encode('utf-8'))
str_fin = [chr(v+2) for v in lst]
print(''.join(str_fin))

To convert a string into a list of character values (numbers), you can use:
s = 'abc'
vals = [ord(c) for c in s]
This results in vals being the list [97, 98, 99].
To convert it back into a string, you can do:
s2 = ''.join(chr(val) for val in vals)
This will give s2 the value 'abc'.
If you prefer to use map rather than comprehensions, you can equivalently do:
vals = list(map(ord, s))
and:
s2 = ''.join(map(chr, vals))
Also, avoid using the name str for a variable, since it will mask the builtin definition of str.

Use ord on the letters to retrieve their decimal ASCII representation, and then chr to convert them back to characters after manipulating the decimal value. Finally use the str.join method with an empty string to piece the list back together into a str:
s = 'abc'
s_list = [ord(let) for let in s]
s_list = [chr(dec + 2) for dec in s_list]
new_s = ''.join(s_list)
print(new_s) # every character is shifted by 2
Calling .encode on the string converts to a bytes string instead, which is likely not what you want. Additionally, you don't want to be using built-ins as the names for variables, because then you will no longer be able to use the built-in keyword in the same scope.

How can I extract words before some string?

I have several strings like this:
mylist = ['pearsapple','grapevinesapple','sinkandapple'...]
I want to parse the parts before apple and then append to a new list:
new = ['pears','grapevines','sinkand']
Is there a way other than finding starting points of 'apple' in each string and then appending before the starting point?

By using slicing in combination with the index method of strings.
>>> [x[:x.index('apple')] for x in mylist]
['pears', 'grapevines', 'sinkand']
You could also use a regular expression
>>> import re
>>> [re.match('(.*?)apple', x).group(1) for x in mylist]
['pears', 'grapevines', 'sinkand']
I don't see why though.

I hope the word apple will be fix (fixed length word), then we can use:
second_list = [item[:-5] for item in mylist]

If some elements in the list don't contain 'apple' at the end of the string, this regex leaves the string untouched:
>>> import re
>>> mylist = ['pearsapple','grapevinesapple','sinkandapple', 'test', 'grappled']
>>> [re.sub('apple$', '', word) for word in mylist]
['pears', 'grapevines', 'sinkand', 'test', 'grappled']

By also using string split and list comprehension
new = [x.split('apple')[0] for x in mylist]
['pears', 'grapevines', 'sinkand']

One way to do it would be to iterate through every string in the list and then use the split() string function.
for word in mylist:
word = word.split("apple")[0]

Sort list of strings alphabetically

So i have a question, how can i sort this list:
['Pera','mela','arancia','UVA']
to be like this:
['arancia','mela','Pera','UVA']
In the exercise it said to use the sorted() function with the cmp argument.

You can easily do that, using the key argument:
my_list = ['Pera','mela','arancia','UVA']
my_list.sort(key=str.lower)
Which will get your lowercases chars first.
This will change the object in-place and my_list will be sorted.
You can use sorted function with the same key argument as well, if you want to have a new list. For example:
my_list = ['Pera','mela','arancia','UVA']
my_sorted_list = sorted(my_list,key=str.lower)
Output will be:
>>> my_list
['Pera','mela','arancia','UVA']
>>> my_sorted_list
['arancia', 'mela', 'Pera', 'UVA']

You need to sort your elements based lowercase representation of the strings:
sorted(['Pera','mela','arancia','UVA'], key=str.lower)
this will output:
['arancia', 'mela', 'Pera', 'UVA']

Use sorted() with a key.
>>> mc = ['Pera','mela','arancia','UVA']
>>> sorted(mc, key=str.lower)
['arancia', 'mela', 'Pera', 'UVA']

This will help you:
>>> words = ['Pera','mela','arancia','UVA']
>>> sorted(words)
['Pera', 'UVA', 'arancia', 'mela']
>>> sorted(words, key=str.swapcase)
['arancia', 'mela', 'Pera', 'UVA']
Hope this helps

create new string by adding characters of smaller strings one after another

You have strings
"aaaaa"
"bbbbb"
"ccccc"
"ddddd"
Now you want to generate the string:
"abcdabcdabcdabcdabcd"
How would be the fastest way to do it?
PS. This is a very simplified example. I really need to generate the new string from the existing smaller strings.

If the strings are of equal length you can use zip:
result = ''.join(map(''.join, zip(*strings)))

Use izip_longest from the itertools library, and the flatten recipe from the same library.
from itertools import izip_longest, chain
def flatten(listOfLists):
"Flatten one level of nesting"
return chain.from_iterable(listOfLists)
result = ''.join(flatten(izip_longest("aaaaa", "bbbbb", "ccc", "dddd", fillvalue='')))

string1 = "aaaaaaaaa"
string2 = "bbbbbbbbb"
string3 = "ccccccccc"
string4 = "ddddddddd"
new_string = ""
for i in range(0,len(string1)):
new_string = new_string+string1[i]+string2[i]+string3[i]+string4[i]
print (new_string)
Results:
abcdabcdabcdabcdabcdabcdabcdabcdabcd