I am struggling with trying to see what is wrong with my code. I am new to python.
import os
uniqueWorms = set()
logLineList = []
with open("redhat.txt", 'r') as logFile:
for eachLine in logFile:
logLineList.append(eachLine.split())
for eachColumn in logLineList:
if 'worm' in eachColumn.lower():
uniqueWorms.append()
print (uniqueWorms)
eachLine.split() returns a list of words. When you append this to logLineList, it becomes a 2-dimensional list of lists.
Then when you iterate over it, eachColumn is a list, not a single column.
If you want logLineList to be a list of words, use
logLineList += eachLine.split()
instead of
logLineList.append(eachLine.split())
Finally, uniqueWorms.append() should be uniqueWOrms.append(eachColumn). And print(uniqueWorms) should be outside the loop, so you just see the final result, not every time a worm is added.
Related
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
Using list comprehension to extract all the tweets in the tweets_and_more list and save them in another list called tweets. Print the length of the resulting list. Also, print the first three elements of the new list.
start_tag = "<tweet>"
def extract_tweet(data):
start = data.index("<tweet>")+len(start_tag)
end = data.index("</tweet>")
seq = data[start:end]
return seq
This is my extract tweet function
My list with all the tweets line by line is saved in a list called "tweets_and_more"
I'm having a hard time figuring out how to use List compression to do this. I can do it using a loop I guess but any help with explanation would be great.
The for loop code is placed at the end
tweets = [extract_tweet(data) for data in tweets_and_more]
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
ladder have around 15000 elements, this code snippet performed in 5-8sec, is there any way to do it faster? I try do it without checking for duplicate and without creating accs list and time was down to 2-3sec, but I don't need duplicate in csv file.
I work in python 2.7.9
accs =[]
with codecs.open('test.csv','w', encoding="UTF-8") as out:
row = ''
for element in ladder:
if element['account']['name'] not in accs:
accs.append(element['account']['name'])
row += element['account']['name']
if 'twitch' in element['account']:
row += "," + element['account']['twitch']['name'] + ","
else:
row += ",,"
row += str(element['account']['challenges']['total']) + "\n"
out.write(row)
seen = set()
results = []
for user in ladder:
acc = user['account']
name = acc['name']
if name not in seen:
seen.add(name)
twitch_name = acc['twitch']['name'] if "twitch" in acc else ''
challenges = acc['challenges']['total']
results.append("%s,%s,%d" % (name, twitch_name, challenges))
with codecs.open('test.csv','w', encoding="UTF-8") as out:
out.write("\n".join(results))
You can’t do much about the loop, since you need to go through every element in ladder after all. However, you can improve this membership test:
if element['account']['name'] not in accs:
Since accs is a list, this will essentially loop through all items of accs and check if the name is in there. And you loop for every element in ladder, so this can easily become very inefficient.
Instead, use a set instead of a list for accs as this will give you a constant membership lookup. So you reduce your algorithm from a quadratic complexity to a linear complexity. For that, just use accs = set() and change your code to use accs.add() instead of append.
Another issue is that you are doing string concatenation. Every time you do someString + "something" you are throwing away that string object and create a new one. This can become inefficient for a high number of operations too. Instead, use a list here to collect all the elements you want to write, and then join them:
row = []
row.append(element['account']['name'])
if 'twitch' in element['account']:
row.append(element['account']['twitch']['name'])
else:
row.append('')
row.append(str(element['account']['challenges']['total']))
out.write(','.join(row))
out.write('\n')
Alternatively, since you are writing to a file anyway, you could just call out.write multiple times with each string part.
Finally, you could also look into the csv module if you are interested in writing out CSV data.
I am new to python and am trying to access a single specific element in a list of lists.
I have tried:
line_list[2][0]
this one isn't right as its a tuple and the list only accepts integers.
line_list[(2, 0)]
line_list[2, 0]
This is probably really obvious but I just can't see it.
def rpd_truncate(map_ref):
#Munipulate string in order to get the reference value
with open (map_ref, "r") as reference:
line_list = []
for line in reference:
word_list = []
word_list.append(line[:-1].split("\t\t"))
line_list.append(word_list)
print line_list[2][0]
I get the exact same as if I used line_list[2]:
['Page_0', '0x00000000', '0x002DF8CD']
actually split will return a list
more over you don't require word_list variable
for line in reference:
line_list.append(line[:-1].split("\t\t"))
print line_list[2][0]