Develop Reference

add str to start of each row value

I have a pandas dataframe
df = pd.DataFrame({'num_legs': [1, 34, 34, 104 , 6542, 6542 , 48383]})
I want to append a str before each row`s value.
The str is ZZ00000
The catch is that the row data must always = 7 characters in total
so the desired output will be
df = num_legs
0 ZZ00001
1 ZZ00034
2 ZZ00034
3 ZZ00104
4 ZZ06542
5 ZZ06542
6 ZZ48383
As the column is of type int I was thinking of changing to a str type and then possibly using regex and some str manipulation to achieve my desired outcome..
Is there a more streamlined way possibly using a function with pandas?

Use
df['num_legs'] = "ZZ" + df['num_legs'].astype(str).str.rjust(5, "0")

You could use string concatenation here:
df["num_legs"] = 'ZZ' + ('00000' + str(df["num_legs"]))[-5:]
The idea here is that, given a num_legs integer value of say 6542, we first form the following string:
000006542
Then we retain the right 5 characters, leaving 06542.

You could also pad using the following:
'ZZ' + df['num_legs'].astype(str).str.pad(width=5, side='left', fillchar='0')
Here you pad your current number (converted to string) on the left with zeros up to a width of 5 and conctatenate that to your 'ZZ' string.

Use pythons .zfill()
df['num_legs']='zz'+df['num_legs'].astype(str).str.zfill(7)

You could try this - using a regex, and a for loop: for strings, for loops are more efficient, usually, than pandas String methods :
import re
variable = "ZZ00000"
df["new_val"] = [re.sub("\d" + f"{{{len(num)}}}$", num, variable)
for num in df.num_legs.astype(str)]
df
num_legs new_val
0 1 ZZ00001
1 34 ZZ00034
2 34 ZZ00034
3 104 ZZ00104
4 6542 ZZ06542
5 6542 ZZ06542
6 48383 ZZ48383

out = []
for nl in df["num_legs"]:
out.append(f'ZZ{nl:05d}')
The rest is up to your output manipulation

How to calculate the verification digit of the Tax ID in the country of Paraguay (calcular digito verificador del RUC)

In the country of Paraguay (South America) each taxpayer has a Tax ID (called RUC: Registro Único del Contribuyente) assigned by the government (Ministerio de Hacienda, Secretaría de Tributación).
This RUC is a number followed by a verification digit (dígito verificador), for example 123456-0. The government tells you the verification digit when you request your RUC.
Is there a way for me to calculate the verification digit based on the RUC? Is it a known formula?
In my case, I have a database of suppliers and customers, collected over the years by several employees of the company.
Now I need to run checks to see if all the RUCs were entered correctly or if there are typing mistakes.
My preference would be a Python solution, but I'll take whatever solutions I get to point me in the right direction.
Edit: This is a self-answer to share knowledge that took me hours/days to find. I marked this question as "answer your own question" (don't know if that changes anything).

The verification digit of the RUC is calculated using formula very similar (but not equal) to a method called Modulo 11; that is at least the info I got reading the following tech sites (content is in Spanish):
https://www.yoelprogramador.com/funncion-para-calcular-el-digito-verificador-del-ruc/
http://groovypy.wikidot.com/blog:02
https://es.wikipedia.org/wiki/C%C3%B3digo_de_control#M.C3.B3dulo_11
I analyzed the solutions provided in the mentioned pages and ran my own tests against a list of RUCs and their known verification digits, which led me to a final formula that returns the expected output, but which is DIFFERENT from the solutions in the mentioned links.
The final formula I got to calculate the verification digit of the RUC is shown in this example (80009735-1):
Multiply each digit of the RUC (without considering the verification digit) by a factor based on the position of the digit within the RUC (starting from the right side of the RUC) and sum all the results of these multiplications:
RUC: 8 0 0 0 9 7 3 5
Position: 7 6 5 4 3 2 1 0
Multiplications: 8x(7+2) 0x(6+2) 0x(5+2) 0x(4+2) 9x(3+2) 7x(2+2) 3x(1+2) 5x(0+2)
Results: 72 0 0 0 45 28 9 10
Sum of results: 164
Divide the sum by 11 and use the remainder of the division to determine the verification digit:
If the remainder is greater than 1, the the verification digit is 11 - remainder
If the remainder is 0 or 1, the the verification digit is 0
In out example:
Sum of results: 164
Division: 164 / 11 ==> quotient 14, remainder 10
Verification digit: 11 - 10 ==> 1
Here is my Python version of the formula:
def calculate_dv_of_ruc(input_str):
# assure that we have a string
if not isinstance(input_str, str):
input_str = str(input_str)
# try to convert to 'int' to validate that it contains only digits.
# I suspect that this is faster than checking each char independently
int(input_str)
the_sum = 0
for i, c in enumerate(reversed(input_str)):
the_sum += (i + 2) * int(c)
base = 11
_, rem = divmod(the_sum, base)
if rem > 1:
dv = base - rem
else:
dv = 0
return dv
Testing this function it returns the expected results, raising errors when the input has other characters than digits:
>>> calculate_dv_of_ruc(80009735)
1
>>> calculate_dv_of_ruc('80009735')
1
>>> calculate_dv_of_ruc('80009735A')
Traceback (most recent call last):
File "<input>", line 1, in <module>
File "<input>", line 8, in calculate_dv_of_ruc
ValueError: invalid literal for int() with base 10: '80009735A'

regex error: bad character range 8-1 at position 6

I am trying to map a new value based on characters in a column. These are digits stored as a string.
If the value of the first character is 1 and the second character is 2-10, then label this as "Lost" etc.
print(x[['Segment']].head(15))
Segment
0 12
1 12
2 22
3 14
4 54
5 12
6 12
7 56
8 12
9 12
10 22
11 12
12 310
13 22
14 53
The mapping I will use:
segt_map = {
r'[4-5][8-10]': 'Champion',
r'[4-5][4-7]': 'Loyal',
r'[4-5][2-3]': 'Recent',
r'3[6-10]': 'High Potential',
r'3[2-5]': 'Need Nurturing',
r'2[6-10]': 'Cannot Lose',
r'2[2-5]': 'At Risk',
r'1[2-10]': 'Lost',
}
And trying to implement it:
x['Label'] = x['Segment'].replace(segt_map, regex=True)
error: bad character range 8-1 at position 6
I am not sure what my error is, and I've checked the related questions, they're not similar to mine. I looked at position 6, you can see it above.. I can't find a range 8-1 ? So what is happening here?
The full error trace is quite long, but if it's needed I can post it.

Character classes match on characters instead of numbers. [8-10] means [8-1] or [0]. You want this:
segt_map = {
r'[4-5]([8-9]|10)': 'Champion',
r'[4-5][4-7]': 'Loyal',
r'[4-5][2-3]': 'Recent',
r'3([6-9]|10)': 'High Potential',
r'3[2-5]': 'Need Nurturing',
r'2([6-9]|10)': 'Cannot Lose',
r'2[2-5]': 'At Risk',
r'1([2-9]|10)': 'Lost',
}

You try to use regex to detect number ranges, whereas it's a tool for processing text that knows nothing about numbers. You cannot use range 8-10, because ranges are for characters. It's ok to use [1-9], because it's about chars, but[1-10] is incorrect. Instead you should parse text ro numbers and then compare them to rewuired ranges.

Cannot understand specific Python 3 code

While trying to solve a problem on Hackerank(The Problem). I checked the solutions of people who have solved this problem. Here is the solution from one person-
n = input()
l = []
for _ in range(n):
s = raw_input().split()
cmd = s[0]
args = s[1:]
if cmd !="print":
cmd += "("+ ",".join(args) +")"
eval("l."+cmd)
else:
print l
I cannot understand line 8 and line 9, can someone explain me these lines? Can i write line 8 and line 9 in Python 3 as i'm learning Python 3 ? How ?

Basically, cmd is constructed by appending the command (say "insert"), to the operands. This cmd forms a correct python expression (for example l.insert(0,5), to insert 5 at index 0, in list l).
Here, l. is hardcoded(start of line 9), cmd is initialized in cmd = s[0], and operands are added in line 8.
eval(str) evaluates the command str, in string format, as if it were in a command line.

It would be nice to just include the problem :) . The input is a (text) file as below:
Sample Input 0
12
insert 0 5
insert 1 10
insert 0 6
print
remove 6
append 9
append 1
sort
print
pop
reverse
print
And the expected output for a correct answer is as below:
Sample Output 0
[6, 5, 10]
[1, 5, 9, 10]
[9, 5, 1]
Before looking at the answer you quoted it would be good to read about eval; it takes the argument provided and tries to run it as a python expression using the global and local namespace. So in this case it needs only the local for the "l"-list and "cmd"-tuple.
What is happening is the following:
Empty list l is created.
The "command" (cmd) single-value list is parsed from the line by slicing (cmd = s[0]), since every line starts with or only has a list method
The other arguments are placed in args
Line 8 (as asked): These other arguments are then joined in a string tuple. So "insert 0 5" gives "insert" for l and "(0, 5)" for cmd
Line 8 continued (as asked): cmd is then combined with args using string concatenation (read here for a good and bad example) resulting in "insert(0,5)" as value for cmd
Line 9 (as asked): the eval parameter is yet another string concatenation yielding "l.insert(0,5)" as final expression to be interpreted. Which then inserts integer value 5 on spot 0 in list l (pushing forward any other values already in l)
Hope it helps, keep on trucking!

Unable to index into a list

I am writing a program to read the output of another program, read it line by line and put it in a list.
#!/usr/bin/python
import subprocess
def RECEIVE(COMMAND):
PROCESS = subprocess.Popen(COMMAND, stdout=subprocess.PIPE)
LINES = iter(PROCESS.stdout.readline, "")
for LINE in LINES:
RECARR = LINE.split()
print RECARR[14]
RECEIVE(["receivetest","-f=/dev/pcan32"])
The output from the receivetest program is:
19327481.401 receivetest: m s 0x0000000663 8 2f 00 42 02 00 e4 8a 8a
19327481.860 receivetest: m s 0x000000069e 8 00 1f 5e 28 34 83 59 1a
it is a constant stream of messages. When split, the list has a range of 14 because after splitting, to make sure, I used:
print len(RECARR)
This gave me an output of 14.
but whenever I try to print the last element:
print RECARR[14]
I get the following error:
file "./cancheck.py", line 10, in RECEIVE
print RECARR[14]
IndexError: list index out of range
This is caused by some erronious text that is printed at the top of the list, so I need some way of making sure that the program only reads in lines that start with
1234567.123
/^(.......\.\d{1,3}) (.*)$/
Any ideas?

Based on the sample data you provided, the length of RECARR is always 14.
14 is the size of the list, not the maximum index. To get the final element of the array, you can try RECARR[13] for this list, or RECARR[-1] in general.
The reason for this is that in Python, as in most programming languages, array indices are zero-based: the first element is accessed with RECARR[0], the second with RECARR[1], and so on. So, the 14th element (or the last one, in your case) would be accessed with RECARR[13].
So, your for loop would look something like this:
for LINE in LINES:
RECARR = LINE.split()
print RECARR[13] # or RECARR[-1]

Right everyone, it's a terrible workaround but I fixed the issue by working out that the only lines with exactly 14 elements are the lines I need so I fixed it by using the following
for LINE in LINES:
RECARR = LINE.split()
if(len(RECARR) == 14):
#do stuff

List indexes start from 0 and not 1. So
print RECARR[1]
prints the 2nd element and not the first. Thus to print the last element you have to use print RECARR[13] or negative index print RECARR[-1].
The lists in python can be depicted as
As you can see the last element can be accessed using either -1 or length of the list -1
An easier way to gauge the ranges is to put the indices before the cell. (Courtesy - Aristide)
+---+---+---+---+---+---+
| P | y | t | h | o | n |
+---+---+---+---+---+---+
0 1 2 3 4 5 6
-6 -5 -4 -3 -2 -1

You could have also done something similar to this:
try:
print RECARR[13]
except IndexError:
pass
This way you can easily handle the lines that are not long enough as well.