How to concatenate string to specific format using Python?

Question:

Good morning,

I am a new python user and I am trying to replicate the following excel function in python:

=concatenate(left(H2,search(" ",H2,1)-1),if(LEN(right(H2,LEN(H2)-search(" ",H2,1)))=2,"00",if(LEN(right(H2,LEN(H2)-search(" ",H2,1)))=3,"0","")),right(H2,LEN(H2)-search(" ",H2,1)),"-KG")

I am trying to convert Highways into a format my computer program(arcgis) reads, basically turn my csv column format from:

enter image description here

the right column into the left column, so basically keep the two letter in the front (ex. US,FM,SL, CR) adding zeros in front making it a 4 digit highway always and adding "-KG" to the end.

thanks

Asked By: Kicker_777

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Answers:

If using Python lists such as

arr = ['US 90', 'FM 1436', 'SL 305', 'US 277', 'FM 1589', 'SL 480']

do

result = [f'{s}{n:0>4}-KG' for s, n in map(str.split, arr)]

which gives

['US0090-KG', 'FM1436-KG', 'SL0305-KG', 'US0277-KG', 'FM1589-KG', 'SL0480-KG']
Answered By: Steven Rumbalski

This would look something like:

import csv

with open("yourfile.csv") as fp:
    reader = csv.reader(fp, delimiter=" ")
    output = [[' '.join(row),f'{row[0]}{row[1].zfill(4)}-KG'] for row in reader]

print(output)

[['US 50', 'US0050-KG'], ['CA 987', 'CA0987-KG'], ['IL 4', 'IL0004-KG']]

The zfill() function will left pad a string with the 0 character to make the string the length of the argument (4 in this case)

If you are wanting to write that list of lists back out, check out this answer

Answered By: JNevill

This function converts the old highway format into your desired format:

def convert(highway):
    split = highway.split()
    prefix = split[0]
    number = split[1]
    return prefix + (4 - len(number))*"0" + number + "-KG"

If you’re working with a pandas dataframe, this converts your whole column:

import pandas as pd

highways = ["US 90", "FM 1436", "SL 305", "US 277", "FM 1689", "SL 480"]

df = pd.DataFrame(highways, columns=["Highways"])
df["Highways"] = df["Highways"].transform(convert, axis=0)

print(df)

Out:

    Highways
0  US0090-KG
1  FM1436-KG
2  SL0305-KG
3  US0277-KG
4  FM1689-KG
5  SL0480-KG

Or if you’re using a list:

highways = ["US 90", "FM 1436", "SL 305", "US 277", "FM 1689", "SL 480"]
new_highways = [convert(highway) for highway in highways]
print(new_highways)

Out:

['US0090-KG', 'FM1436-KG', 'SL0305-KG', 'US0277-KG', 'FM1689-KG', 'SL0480-KG']
Answered By: Timo

You can try using the string format instead of zfill:
Here’s an example it’s not for csv but in arcgis I think you wanted to use it in the atributte calculator.

List= ['US 90','FM 1436','SL 305']

for elem in List:
    Splited = str(elem).split(' ')
    Concatenated_elem = str(Splited[0]) + f'{int(Splited[1]):04d}' +'-KG'
    print(Concatenated_elem)

The result in this case is a print:

US0090-KG
FM1436-KG
SL0305-KG

Hope it helps,

Answered By: SSD93
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