Develop Reference

Editing String Objects in a List in Python

I have read in data from a basic txt file. The data is time and date in this form "DD/HHMM" (meteorological date and time data). I have read this data into a list: time[]. It prints out as you would imagine like so: ['15/1056', '15/0956', '15/0856', .........]. Is there a way to alter the list so that it ends up just having the time, basically removing the date and the forward slash, like so: ['1056', '0956', '0856',.........]? I have already tried list.split but thats not how that works I don't think. Thanks.

I'm still learning myself and I haven't touched python in sometime, BUT, my solution if you really need one:
myList = ['15/1056', '15/0956', '15/0856']
newList = []
for x in mylist:
newList.append(x.split("/")[1])
# splits at '/'
# returns ["15", "1056"]
# then appends w/e is at index 1
print(newList) # for verification

Parsing and arranging text in python

I'm having some trouble figuring out the best implementation
I have data in file in this format:
|serial #|machine_name|machine_owner|
If a machine_owner has multiple machines, I'd like the machines displayed in a comma separated list in the field. so that.
|1234|Fred Flinstone|mach1|
|5678|Barney Rubble|mach2|
|1313|Barney Rubble|mach3|
|3838|Barney Rubble|mach4|
|1212|Betty Rubble|mach5|
Looks like this:
|Fred Flinstone|mach1|
|Barney Rubble|mach2,mach3,mach4|
|Betty Rubble|mach5|
Any hints on how to approach this would be appreciated.

You can use dict as temporary container to group by name and then print it in desired format:
import re
s = """|1234|Fred Flinstone|mach1|
|5678|Barney Rubble|mach2|
|1313|Barney Rubble||mach3|
|3838|Barney Rubble||mach4|
|1212|Betty Rubble|mach5|"""
results = {}
for line in s.splitlines():
_, name, mach = re.split(r"\|+", line.strip("|"))
if name in results:
results[name].append(mach)
else:
results[name] = [mach]
for name, mach in results.items():
print(f"|{name}|{','.join(mach)}|")

You need to store all the machines names in a list. And every time you want to append a machine name, you run a function to make sure that the name is not already in the list, so that it will not put it again in the list.
After storing them in an array called data. Iterate over the names. And use this function:
data[i] .append( [ ] )
To add a list after each machine name stored in the i'th place.
Once your done, iterate over the names and find them in in the file, then append the owner.
All of this can be done in 2 steps.

Count and flag duplicates in a column in a csv

this type of question has been asked many times. So apologies; I have searched hard to get an answer - but have not found anything that is close enough to my needs (and I am not sufficiently advanced (I am a total newbie) to customize an existing answer). So thanks in advance for any help.
Here's my query:
I have 30 or so csv files and each contains between 500 and 15,000 rows.
Within each of them (in the 1st column) - are rows of alphabetical IDs (some contain underscores and some also have numbers).
I don't care about the unique IDs - but I would like to identify the duplicate IDs and the number of times they appear in all the different csv files.
Ideally I'd like the output for each duped ID to appear in a new csv file and be listed in 2 columns ("ID", "times_seen")
It may be that I need to compile just 1 csv with all the IDs for your code to run properly - so please let me know if I need to do that
I am using python 2.7 (a crawling script that I run needs this version, apparently).
Thanks again

It seems the most easy way to achieve want you want would make use of dictionaries.
import csv
import os
# Assuming all your csv are in a single directory we will iterate on the
# files in this directory, selecting only those ending with .csv
# to list files in the directory we will use the walk function in the
# os module. os.walk(path_to_dir) returns a generator (a lazy iterator)
# this generator generates tuples of the form root_directory,
# list_of_directories, list_of_files.
# So: declare the generator
file_generator = os.walk("/path/to/csv/dir")
# get the first values, as we won't recurse in subdirectories, we
# only ned this one
root_dir, list_of_dir, list_of_files = file_generator.next()
# Now, we only keep the files ending with .csv. Let me break that down
csv_list = []
for f in list_of_files:
if f.endswith(".csv"):
csv_list.append(f)
# That's what was contained in the line
# csv_list = [f for _, _, f in os.walk("/path/to/csv/dir").next() if f.endswith(".csv")]
# The dictionary (key value map) that will contain the id count.
ref_count = {}
# We loop on all the csv filenames...
for csv_file in csv_list:
# open the files in read mode
with open(csv_file, "r") as _:
# build a csv reader around the file
csv_reader = csv.reader(_)
# loop on all the lines of the file, transformed to lists by the
# csv reader
for row in csv_reader:
# If we haven't encountered this id yet, create
# the corresponding entry in the dictionary.
if not row[0] in ref_count:
ref_count[row[0]] = 0
# increment the number of occurrences associated with
# this id
ref_count[row[0]]+=1
# now write to csv output
with open("youroutput.csv", "w") as _:
writer = csv.writer(_)
for k, v in ref_count.iteritems():
# as requested we only take duplicates
if v > 1:
# use the writer to write the list to the file
# the delimiters will be added by it.
writer.writerow([k, v])
You may need to tweek a little csv reader and writer options to fit your needs but this should do the trick. You'll find the documentation here https://docs.python.org/2/library/csv.html. I haven't tested it though. Correcting the little mistakes that may have occurred is left as a practicing exercise :).

That's rather easy to achieve. It would look something like:
import os
# Set to what kind of separator you have. '\t' for TAB
delimiter = ','
# Dictionary to keep count of ids
ids = {}
# Iterate over files in a dir
for in_file in os.listdir(os.curdir):
# Check whether it is csv file (dummy way but it shall work for you)
if in_file.endswith('.csv'):
with open(in_file, 'r') as ifile:
for line in ifile:
my_id = line.strip().split(delimiter)[0]
# If id does not exist in a dict = set count to 0
if my_id not in ids:
ids[my_id] = 0
# Increment the count
ids[my_id] += 1
# saves ids and counts to a file
with open('ids_counts.csv', 'w') as ofile:
for key, val in ids.iteritems():
# write down counts to a file using same column delimiter
ofile.write('{}{}{}\n'.format(key, delimiter, value))

Check out the pandas package. You can read an write csv files quite easily with it.
http://pandas.pydata.org/pandas-docs/stable/10min.html#csv
Then, when having the csv-content as a dataframe you convert it with the as_matrix function.
Use the answers to this question to get the duplicates as a list.
Find and list duplicates in a list?
I hope this helps

As you are a newbie, Ill try to give some directions instead of posting an answer. Mainly because this is not a "code this for me" platform.
Python has a library called csv, that allows to read data from CSV files (Boom!, surprised?). This library allows you to read the file. Start by reading the file (preferably an example file that you create with just 10 or so rows and then increase the amount of rows or use a for loop to iterate over different files). The examples in the bottom of the page that I linked will help you printing this info.
As you will see, the output you get from this library is a list with all the elements of each row. Your next step should be extracting just the ID that you are interested in.
Next logical step is counting the amount of appearances. There is also a class from the standard library called counter. They have a method called update that you can use as follows:
from collections import Counter
c = Counter()
c.update(['safddsfasdf'])
c # Counter({'safddsfasdf': 1})
c['safddsfasdf'] # 1
c.update(['safddsfasdf'])
c # Counter({'safddsfasdf': 2})
c['safddsfasdf'] # 2
c.update(['fdf'])
c # Counter({'safddsfasdf': 2, 'fdf': 1})
c['fdf'] # 1
So basically you will have to pass it a list with the elements you want to count (you could have more than 1 id in the list, for exampling reading 10 IDs before inserting them, for improved efficiency, but remember not constructing a thousands of elements list if you are seeking good memory behaviour).
If you try this and get into some trouble come back and we will help further.
Edit
Spoiler alert: I decided to give a full answer to the problem, please avoid it if you want to find your own solution and learn Python in the progress.
# The csv module will help us read and write to the files
from csv import reader, writer
# The collections module has a useful type called Counter that fulfills our needs
from collections import Counter
# Getting the names/paths of the files is not this question goal,
# so I'll just have them in a list
files = [
"file_1.csv",
"file_2.csv",
]
# The output file name/path will also be stored in a variable
output = "output.csv"
# We create the item that is gonna count for us
appearances = Counter()
# Now we will loop each file
for file in files:
# We open the file in reading mode and get a handle
with open(file, "r") as file_h:
# We create a csv parser from the handle
file_reader = reader(file_h)
# Here you may need to do something if your first row is a header
# We loop over all the rows
for row in file_reader:
# We insert the id into the counter
appearances.update(row[:1])
# row[:1] will get explained afterwards, it is the first column of the row in list form
# Now we will open/create the output file and get a handle
with open(output, "w") as file_h:
# We create a csv parser for the handle, this time to write
file_writer = writer(file_h)
# If you want to insert a header to the output file this is the place
# We loop through our Counter object to write them:
# here we have different options, if you want them sorted
# by number of appearances Counter.most_common() is your friend,
# if you dont care about the order you can use the Counter object
# as if it was a normal dict
# Option 1: ordered
for id_and_times in apearances.most_common():
# id_and_times is a tuple with the id and the times it appears,
# so we check the second element (they start at 0)
if id_and_times[1] == 1:
# As they are ordered, we can stop the loop when we reach
# the first 1 to finish the earliest possible.
break
# As we have ended the loop if it appears once,
# only duplicate IDs will reach to this point
file_writer.writerow(id_and_times)
# Option 2: unordered
for id_and_times in apearances.iteritems():
# This time we can not stop the loop as they are unordered,
# so we must check them all
if id_and_times[1] > 1:
file_writer.writerow(id_and_times)
I offered 2 options, printing them ordered (based on Counter.most_common() doc) and unoredered (based on normal dict method dict.iteritems()). Choose one. From a speed point of view I'm not sure which one would be faster, as one first needs to order the Counter but also stops looping when finding the first element non-duplicated while the second doesn't need to order the elements but needs to loop every ID. The speed will probably be dependant on your data.
About the row[:1] thingy:
row is a list
You can get a subset of a list telling the initial and final positions
In this case the initial position is omited, so it defaults to the start
The final position is 1, so just the first element gets selected
So the output is another list with just the first element
row[:1] == [row[0]] They have the same output, getting a sublist of only the same element is the same that constructing a new list with only the first element

Most efficient way to merge two partially inclusive lists of files and their properties

I have a system that runs a custom cli with a variation of the ls or dir command, returning a list of files and folders in your working directory.
The problem is, I can either run the command with a flag that returns the files and their time stamps (date created and last modified), or one that returns the file and their file sizes. There is no way to get both in a single cli command.
A further complication arises when getting the time stamped list, only some of the files are returned (all files ending in certain prefixes are left out). Neither list is in any particular order.
I wish to create a dictionary that contains all the information for each file in one place. What is the cleanest, most efficient, and most pythonic way to do this?
Quick sample of data:
dir -time gives a list of 506 elements. Only (but not all) files ending in .ts have timestamps. Some files show in the list but do not have timestamps, some files (such as anything ending in .index) do not show up in the list at all.
ch20prefix_20_182.ts 2014-10-22 16:06:20 - 2014-10-22 16:08:51
ch21prefix_21_40.ts 2014-10-14 16:15:42 - 2014-10-14 16:16:51
modinfo_sdk1.23b24L
bs780_ntplatency
ch10prefix_10_237.ts 2014-10-27 11:05:10 - 2014-10-27 11:07:33
ch10prefix_10_277.ts 2014-10-30 14:03:51 - 2014-10-30 14:04:24
video1_6_1.ts
ch11prefix_11_179.ts 2014-10-22 14:53:50 - 2014-10-22 14:56:00`
dir -size gives a list of 967 elements. All files are present here, all files have a file size.
ch10prefix_10_340.index 159544
ch2prefix_2_705.ts 75958204
<ts220> 0
ch11prefix_11_148.ts 19877616
ch10prefix_10_310.ts 7373924
ch11prefix_11_111.index 17112
ch11prefix_11_278.index 1368
ch2prefix_2_307.ts 6492580
channelConfig.xml.2HD 18144
ch21prefix_21_220.ts 12893604
ch20prefix_20_128.index 1720
There is some rhyme and reason to the mess that is why some files show up and others don't, why some have timestamps and others don't, but that is largely irrelevant to this question.
My thoughts on how to approach it:
What I want as final output is a dictionary with each key as a file name, and it's value as another dictionary with key/val pairs for Time Created, Time Mod, fileSize. This way one can easily lookup all 3 pieces of information for each file.
The difficult part for me, however, is finding an efficient way of combining the data from each list. The first thing that comes to mind would be to loop through the larger list (file size), and then for each element, check if it is in the smaller list, and if it is (and has a timestamp), add the data. But that is horridly inefficient. Although some files in the larger list I know ahead of time do not have timestamps in the other list, I cannot say that for all files that don't have a timestamp.
The lists are unsorted, but It occurs to me that if they were sorted by file name, that allow for a much faster way of looking up each file from one list in another, but considering the runtime of sorting the lists, it still might not be worth the effort.
So, what would be the most efficient approach here? I am mostly concerned with run-time and readability, but welcome the inclusion of other factors in how I might approach this problem.

It is hard to tell from your question what your desired result is. If you want all files in both lists even if they only appear in one or the other just make one pass through both files and create a dictionary using collections.defaultdict
from collections import defaultdict
d = defaultdict(dict)
with open('fileA.txt') as f:
for line in f:
name, time = line[:24], line[24:]
name, time = name.strip(), time.strip()
time_created, time_modified = time.split(' - ')
d[name]['time_created'] = time_created
d[name]['time_modified'] = time_modified
with open('fileB.txt') as f:
for line in f:
name, size = line[:24], line[24:]
name, size = name.strip(), size.strip()
d[name]['size'] = size
If your final result only includes files that appear in both lists then make one pass over each list constructing separate dictionaries.
dA = defaultdict(dict)
dB = defaultdict(dict)
with open('fileA.txt') as f:
for line in f:
name, time = line[:24], line[24:]
name, time = name.strip(), time.strip()
try:
time_created, time_modified = time.split(' - ')
except ValueError:
time_created, time_modified = '', ''
dA[name]['time_created'] = time_created
dA[name]['time_modified'] = time_modified
with open('fileB.txt') as f:
for line in f:
name, size = line[:24], line[24:]
name, size = name.strip(), size.strip()
dB[name]['size'] = size
Then make a pass over one of those dictionaries creating a third dictionary with common keys.
d = defaultdict(dict)
for k, v in dA.items():
if k in dB:
d[k] = v
d[k].update(dB[k])
Since this is the only answer (so far) with a solution And #Brian C didn't offer one, this MUST be the most efficient.

Sounds like a good use case for Sqlite.
Python has good support for it. Instead of creating a disk file based DB you could use a pure in-memory based database by passing the right arguments. First I'd create a 2 tables - tblFileNTimeStamp (File name (PK), timestamp) and tblFileNSize (File name (PK), filesize). Use the output of the two commands to populate the database and use a join on the primary keys to pick the results you need.

Best way to parse a file with columns that randomly change order before importing it into SQL Server 2008?

I have a file that has columns that look like this:
Column1,Column2,Column3,Column4,Column5,Column6
1,2,3,4,5,6
1,2,3,4,5,6
1,2,3,4,5,6
1,2,3,4,5,6
1,2,3,4,5,6
1,2,3,4,5,6
Column1,Column3,Column2,Column6,Column5,Column4
1,3,2,6,5,4
1,3,2,6,5,4
1,3,2,6,5,4
Column2,Column3,Column4,Column5,Column6,Column1
2,3,4,5,6,1
2,3,4,5,6,1
2,3,4,5,6,1
The columns randomly re-order in the middle of the file, and the only way to know the order is to look at the last set of headers right before the data (Column1,Column2, etc.) (I've also simplified the data so that it's easier to picture. In real life, there is no way to tell data apart as they are all large integer values that could really go into any column)
Obviously this isn't very SQL Server friendly when it comes to using BULK INSERT, so I need to find a way to arrange all of the columns in a consistent order that matches my table's column order in my SQL database. What's the best way to do this? I've heard Python is the language to use, but I have never worked with it. Any suggestions/sample scripts in any language are appreciated.

A solution in python:
I would read line-by-line and look for headers. When I find a header, I use it to figure out the order (somehow). Then I pass that order to itemgetter which will do the magic of reordering elements:
from operator import itemgetter
def header_parse(line,order_dict):
header_info = line.split(',')
indices = [None] * len(header_info)
for i,col_name in enumerate(header_info):
indices[order_dict[col_name]] = i
return indices
def fix(fname,foutname):
with open(fname) as f,open(foutname,'w') as fout:
#Assume first line is a "header" and gives the order to use for the
#rest of the file
line = f.readline()
order_dict = dict((name,i) for i,name in enumerate(line.strip().split(',')))
reorder_magic = itemgetter(*header_parse(line.strip(),order_dict))
for line in f:
if line.startswith('Column'): #somehow determine if this is a "header"
reorder_magic = itemgetter(*header_parse(line.strip(),order_dict))
else:
fout.write(','.join(reorder_magic(line.strip().split(','))) + '\n')
if __name__ == '__main__':
import sys
fix(sys.argv[1],sys.argv[2])
Now you can call it as:
python fixscript.py badfile goodfile

Since you didn't mention a specific problem, I'm going to assume you're having problems coming up with an algorithm.
For each row,
Parse the row into fields.
If it's the first header line,
Output the header.
Create a map of field names to position.
%map = map { $fields[$_] => $_ } 0..$#fields;
Create a map of original positions to new positions.
#map = #map{ #fields };
If it's a header line other than the first,
Update map of original positions to new positions.
#map = #map{ #fields };
If it's not a header line,
Reorder fields.
#fields[ #map ] = #fields;
Output the row.
(Snippets are in Perl.)

This can be fixed easily in two steps:
split file into multiple files when a new header starts
read each file using csv dict reader, sort the keys and re-output rows in correct order
Here is an example how you can ho about it,
def is_header(line):
return line.find('Column') >= 0
def process(lines):
headers = None
for line in lines:
line = line.strip()
if is_header(line):
headers = list(enumerate(line.split(",")))
headers_map = dict(headers)
headers.sort(key=lambda (i,v):headers_map[i])
print ",".join([h for i,h in headers])
continue
values = list(enumerate(line.split(",")))
values.sort(key=lambda (i,v):headers_map[i])
print ",".join([v for i,v in values])
if __name__ == "__main__":
import sys
process(open(sys.argv[1]))
You can also change function is_header to correctly identify header in real cases