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Remove leading zeros in forecast period string

I am needing to format forecast period columns to later merge with another data frame.
Columns of my data frame are:
current_cols = [
'01+11',
'02+10',
'03+09',
'04+08',
'05+07',
'06+06',
'07+05',
'08+04',
'09+03',
'10+02',
'11+01'
]
desired_out = [
'1+11',
'2+10',
'3+9',
'4+8',
'5+7',
'6+6',
'7+5',
'8+4',
'9+3',
'10+2',
'11+1'
]
Originally, I tried to split the list by split('+'), and use lstrip('0') for each element in the list. Then recombine elements within tuple with + in between.
Is there a better approach? I'm having trouble combining elements in tuples back together, with + in between. Help would be much appreciated.

You can use re module for the task:
import re
pat = re.compile(r"\b0+")
out = [pat.sub(r"", s) for s in current_cols]
print(out)
Prints:
[
"1+11",
"2+10",
"3+9",
"4+8",
"5+7",
"6+6",
"7+5",
"8+4",
"9+3",
"10+2",
"11+1",
]

current_cols =['01+11','02+10','03+09','04+08','05+07','06+06','07+05','08+04','09+03','10+02','11+01']
desired_out = []
for item in current_cols:
if item[0] == "0":
item = item[1:]
if "+0" in item:
item = item.replace('+0', '+')
desired_out.append(item)

You can do it with nested comprehensions, conversion to int(), and formatting using an f-string:
current_cols = [
'01+11',
'02+10',
'03+09',
'04+08',
'05+07',
'06+06',
'07+05',
'08+04',
'09+03',
'10+02',
'11+01'
]
desired_out = [
f'{int(a)}+{int(b)}' for (a, b) in [
e.split('+') for e in current_cols
]
]
The code above will set desired_out with:
['1+11', '2+10', '3+9', '4+8', '5+7', '6+6', '7+5', '8+4', '9+3', '10+2', '11+1']
This method is implementing your original thought of splitting each element using the + signal as separator, extracting the leading zeros from each pair element (done with the int() conversion inside the f-string), and combining them back, with a + sign in between (also using the f-string).
The inner comprehension is just walking each element of the list, and splitting them by the + sign. The outer comprehension converts each element of each pair to int() to get rid of the leading zeros.

We want a bunch of map operations here to do the following:
split each element of current_cols on "+":
map(lambda s: s.split("+"), current_cols)
lstrip the "0" out of each element of the resulting lists:
map(lambda l: (x.lstrip("0") for x in l), ...)
join the resulting values on "+":
map("+".join, ...)
Then, we list out the elements of these map operations:
list(
map("+".join,
map(lambda l: (x.lstrip('0') for x in l),
map(lambda s: s.split('+'), current_cols)
)
)
)
which gives:
['1+11',
'2+10',
'3+9',
'4+8',
'5+7',
'6+6',
'7+5',
'8+4',
'9+3',
'10+2',
'11+1']

How to group all the first characters of a string in a list of string , all second character of a string and so on in a list of string in python

a=["cypatlyrm","aolsemone","nueeleuap"]
o/p needed is : canyoupleasetellmeyournamep
I have tried
for i in range(len(a)):
for j in range(len(a)):
res+=a[j][i]
it gives o/p : canyouple
how to get full output ?

You can use itertools.zip_longest with fill value as empty string'' and itertools.chain and the join the result to get what you want.
from itertools import zip_longest, chain
seq = ["cypatlyrm", "aolsemone", "nueeleuap"]
res = ''.join(chain.from_iterable(zip_longest(*seq, fillvalue='')))
print(res)
Output
canyoupleasetellmeyournamep
Using zip_longest makes sure that this also works with cases where the element sizes are not equal. If all elements in the list are guaranteed to be the same length then a normal zip would also work.
If all the elements have the same length then you can use this approach that does not need libraries that have to be imported.
seq = ["cypatlyrm", "aolsemone", "nueeleuap"]
res = ''
for i in range(len(seq[0])):
for j in seq:
res += j[i]
print(res)

Sum all numbers in a list of strings

sorry if this is very noob question, but I have tried to solve this on my own for some time, gave it a few searches (used the "map" function, etc.) and I did not find a solution to this. Maybe it's a small mistake somewhere, but I am new to python and seem to have some sort of tunnel vision.
I have some text (see sample) that has numbers inbetween. I want to extract all numbers with regular expressions into a list and then sum them. I seem to be able to do the extraction, but struggle to convert them to integers and then sum them.
import re
df = ["test 4497 test 6702 test 8454 test",
"7449 test"]
numlist = list()
for line in df:
line = line.rstrip()
numbers = re.findall("[0-9]+", line) # find numbers
if len(numbers) < 1: continue # ignore lines with no numbers, none in this sample
numlist.append(numbers) # create list of numbers
The sum(numlist) returns an error.

You don't need a regex for this. Split the strings in the list, and sum those that are numeric in a comprehension:
sum(sum(int(i) for i in s.split() if i.isnumeric()) for s in df)
# 27102
Or similarly, flatten the resulting lists, and sum once:
from itertools imprt chain
sum(chain.from_iterable((int(i) for i in s.split() if i.isnumeric()) for s in df))
# 27102

This is the source of your problem:
finadall returns a list which you are appending to numlist, a list. So you end up with a list of lists. You should instead do:
numlist.extend(numbers)
So that you end up with a single list of numbers (well, actually string representations of numbers). Then you can convert the strings to integers and sum:
the_sum = sum(int(n) for n in numlist)

Iterate twice over df and append each digit to numlist:
numlist = list()
for item in df:
for word in item.split():
if word.isnumeric():
numlist.append(int(word))
print(numlist)
print(sum(numlist))
Out:
[4497, 6702, 8454, 7449]
27102
You could make a one-liner using list comprehension:
print(sum([int(word) for item in df for word in item.split() if word.isnumeric()]))
>>> 27102

It's as easy as
my_sum = sum(map(int, numbers_list))

Here is an option using map, filter and sum:
First splits the strings at the spaces, filters out the non-numbers, casts the number-strings to int and finally sums them.
# if you want the sum per string in the list
sums = [sum(map(int, filter(str.isnumeric, s.split()))) for s in df]
# [19653, 7449]
# if you simply want the sum of all numbers of all strings
sum(sum(map(int, filter(str.isnumeric, s.split()))) for s in df)
# 27102

How to search through an arry containing strings, and create a new array with only integers

If I have an array that contains only strings, but some of them are numbers, how would I search through the array, determine which strings are actually numbers, and add those numbers to a new array? An example of the array is as follows: [ "Chris" , "90" , "Dave" , "76" ]
I have tried using a for loop to consecutively use isdigit() on each index, and if it is true to add that item to the new array.
scores = []
for i in range(len(name_and_score_split)):
if name_and_score_split[i].isdigit() == True:
scores.append(name_and_score_split[i])
When the above code is ran it tells me list data type does not have the "isdigit" function
edit: iv'e found that my problem is the list is actually a list of lists.

Use a list-comprehension and also utilise the for-each property of Python for rather than iterating over indices:
lst = ["Chris" , "90" , "Dave" , "76"]
scores = [x for x in lst if x.isdigit()]
# ['90', '76']
Alternately, filter your list:
scores = list(filter(lambda x: x.isdigit(), lst))

Assuming what you're trying if for integers you can do something like:
// Taken from and changing float by int.
def is_number(s):
try:
int(s)
return True
except ValueError:
return False
Then you can do
[x for x in name_and_score_split if is_number(x)]

If you want list of int:
s = ["Chris", "90", "Dave", "76"]
e = [int(i) for i in s if i.isdigit()]
print(e)
# OUTPUT: [90, 76]

searching a list of strings for integers

Given the following list of strings:
my_list = ['element0 123 321\n', 'element1 223 32221\n', 'element2 19823 328771\n', ... ]
how can I split each entry into a list of tuples:
[ (123, 321), (223, 32221), (19823, 328771), ... ]
In my other poor attempt, I managed to extract the numbers, but I encountered a problem, the element placeholder also contains a number which this method includes! It also doesn't write to a tuple, rather a list.
numbers = list()
for s in my_list:
for x in s:
if x.isdigit():
numbers.append((x))
numbers

We can first build a regex that identifies positive integers:
from re import compile
INTEGER_REGEX = compile(r'\b\d+\b')
Here \d stands for digit (so 0, 1, etc.), + for one or more, and \b are word boundaries.
We can then use INTEGER_REGEX.findall(some_string) to identify all positive integers from the input. Now the only thing left to do is iterate through the elements of the list, and convert the output of INTEGER_REGEX.findall(..) to a tuple. We can do this with:
output = [tuple(INTEGER_REGEX.findall(l)) for l in my_list]
For your given sample data, this will produce:
>>> [tuple(INTEGER_REGEX.findall(l)) for l in my_list]
[('123', '321'), ('223', '32221'), ('19823', '328771')]
Note that digits that are not separate words will not be matched. For instance the 8 in 'see you l8er' will not be matched, since it is not a word.

your attempts iterates on each char of the string. You have to split the string according to blank. A task that str.split does flawlessly.
Also numbers.append((x)) is numbers.append(x). For a tuple of 1 element, add a comma before the closing parenthese. Even if that doesn't solve it either.
Now, the list seems to contain an id (skipped), then 2 integers as string, so why not splitting, zap the first token, and convert as tuple of integers?
my_list = ['element0 123 321\n', 'element1 223 32221\n', 'element2 19823 328771\n']
result = [tuple(map(int,x.split()[1:])) for x in my_list]
print(result)
gives:
[(123, 321), (223, 32221), (19823, 328771)]