I wrote this function which does the following:
after readling lines from file.It splits and finds the word occurences
through a hash table...for some reason this is quite slow..can some one
help me make it faster...
f = open(filename)
lines = f.readlines()
def create_words(lines):
cnt = 0
spl_set = '[",;<>{}_&?!():-[\.=+*\t\n\r]+'
for content in lines:
words=content.split()
countDict={}
wordlist = []
for w in words:
w=string.lower(w)
if w[-1] in spl_set: w = w[:-1]
if w != '':
if countDict.has_key(w):
countDict[w]=countDict[w]+1
else:
countDict[w]=1
wordlist = countDict.keys()
wordlist.sort()
cnt += 1
if countDict != {}:
for word in wordlist: print (word+' '+
str(countDict[word])+'\n') 19 1938
why reload wordlist and sort it after each word processing ? seems that
it can be done after the for loop. pk******@gmail.com wrote: I wrote this function which does the following: after readling lines from file.It splits and finds the word occurences through a hash table...for some reason this is quite slow..can some one help me make it faster... f = open(filename) lines = f.readlines() def create_words(lines): cnt = 0 spl_set = '[",;<>{}_&?!():-[\.=+*\t\n\r]+' for content in lines: words=content.split() countDict={} wordlist = [] for w in words: w=string.lower(w) if w[-1] in spl_set: w = w[:-1] if w != '': if countDict.has_key(w): countDict[w]=countDict[w]+1 else: countDict[w]=1 wordlist = countDict.keys() wordlist.sort() cnt += 1 if countDict != {}: for word in wordlist: print (word+' '+ str(countDict[word])+'\n')
Actually I create a seperate wordlist for each so called line.Here line
I mean would be a paragraph in future...so I will have to recreate the
wordlist for each loop
Oh sorry indentation was messed here...the
wordlist = countDict.keys()
wordlist.sort()
should be outside the word loop.... now
def create_words(lines):
cnt = 0
spl_set = '[",;<>{}_&?!():-[\.=+*\t\n\r]+'
for content in lines:
words=content.split()
countDict={}
wordlist = []
for w in words:
w=string.lower(w)
if w[-1] in spl_set: w = w[:-1]
if w != '':
if countDict.has_key(w):
countDict[w]=countDict[w]+1
else:
countDict[w]=1
wordlist = countDict.keys()
wordlist.sort()
cnt += 1
if countDict != {}:
for word in wordlist: print (word+' '+
str(countDict[word])+'\n')
ok now this is the correct question I am asking...
don't know your intend so have no idea what it is for. However, you are
doing :
wordlist=contDict.keys()
wordlist.sort()
for every word processed yet you don't use the content of x in anyway
during the loop. Even if you need one fresh snapshot of contDict after
each word, I don't see the need for sorting. seems like wasting cycles
to me.
pkila...@gmail.com wrote: Actually I create a seperate wordlist for each so called line.Here line I mean would be a paragraph in future...so I will have to recreate the wordlist for each loop
You're making a new countDict for each line read from the file... is
that what you meant to do? Or are you trying to count word occurrences
across the whole file?
--
In general, any time string manipulation is going slowly, ask yourself,
"Can I use the re module for this?"
# disclaimer: untested code. probably contains typos
import re
word_finder = re.compile('[a-z0-9_]+', re.I)
def count_words (string, word_finder = word_finder): # avoid global
lookups
countDict = {}
for match in word_finder.finditer(string):
word = match.group(0)
countDict[word] = countDict.get(word,0) + 1
return countDict
f = open(filename)
for i, line in enumerate(f.xreadlines()):
countDict = count_words(line)
print "Line %s" % i
for word in sorted(countDict.keys()):
print " %s %s" % (word, countDict[word])
f.close()
This can be faster, it avoids doing the same things more times:
from string import maketrans, ascii_lowercase, ascii_uppercase
def create_words(afile):
stripper = """'[",;<>{}_&?!():[]\.=+-*\t\n\r^%0123456789/"""
mapper = maketrans(stripper + ascii_uppercase,
" "*len(stripper) + ascii_lowercase)
countDict = {}
for line in afile:
for w in line.translate(mapper).split():
if w:
if w in countDict:
countDict[w] += 1
else:
countDict[w] = 1
word_freq = countDict.items()
word_freq.sort()
for word, freq in word_freq:
print word, freq
create_words(file("test.txt"))
If you can load the whole file in memory then it can be made a little
faster...
Bear hugs,
bearophile pk******@gmail.com wrote: I wrote this function which does the following: after readling lines from file.It splits and finds the word occurences through a hash table...for some reason this is quite slow..can some one help me make it faster... f = open(filename) lines = f.readlines() def create_words(lines): cnt = 0 spl_set = '[",;<>{}_&?!():-[\.=+*\t\n\r]+' for content in lines: words=content.split() countDict={} wordlist = [] for w in words: w=string.lower(w) if w[-1] in spl_set: w = w[:-1] if w != '': if countDict.has_key(w): countDict[w]=countDict[w]+1 else: countDict[w]=1 wordlist = countDict.keys() wordlist.sort() cnt += 1 if countDict != {}: for word in wordlist: print (word+' '+ str(countDict[word])+'\n')
The way this is written you create a new countDict object
for every line of the file, it's not clear that this is
what you meant to do.
Also you are sorting wordlist for every line, not just
the entire file because it is inside the loop that is
processing lines.
Some extra work by testing for empty dictionary:
wordlist=countDict.keys()
then
if countdict != {}:
for word in wordlist:
if countDict is empty then wordlist will be empty so testing
for it is unnecessary.
Incrementing cnt, but never using it.
I don't think spl_set will do what you want, but I haven't modified
it. To split on all those characters you are going to need to
use regular expressions not split.
Modified code:
def create_words(lines):
spl_set = '[",;<>{}_&?!():-[\.=+*\t\n\r]+'
countDict={}
for content in lines:
words=content.split()
for w in words:
w=w.lower()
if w[-1] in spl_set: w = w[:-1]
if w:
if countDict.has_key(w):
countDict[w]=countDict[w]+1
else:
countDict[w]=1
return countDict
import time
filename=r'C:\cygwin\usr\share\vim\vim63\doc\versi on5.txt'
f = open(filename)
lines = f.readlines()
start_time=time.time()
countDict=create_words(lines)
stop_time=time.time()
elapsed_time=stop_time-start_time
wordlist = countDict.keys()
wordlist.sort()
for word in wordlist:
print "word=%s count=%i" % (word, countDict[word])
print "Elapsed time in create_words function=%.2f seconds" % elapsed_time
I ran this against a 551K text file and it runs in 0.11 seconds
on my machine (3.0Ghz P4).
Larry Bates
ok this sounds much better..could you tell me what to do if I want to
leave characters like @ in words.So I would like to consider this as a
part of word
The word_finder regular expression defines what will be considered a
word.
"[a-z0-9_]" means "match a single character from the set {a through z,
0 through 9, underscore}".
The + means "match as many as you can, minimum of one"
To match @ as well, add it to the set of characters to match:
word_finder = re.compile('[a-z0-9_@]+', re.I)
The re.I flag makes the expression case insensitive.
See the documentation for re for more information.
Also--- It looks like I forgot to lowercase matched words. The line
word = match.group(0)
should read:
word = match.group(0).lower()
Lonnie Princehouse wrote: "[a-z0-9_]" means "match a single character from the set {a through z, 0 through 9, underscore}".
"\w" should be a bit faster; it's equivalent to "[a-zA-Z0-9_]" (unless you
specify otherwise using the locale or unicode flags), but is handled more
efficiently by the RE engine.
(you can use \w in sets too, so you can do e.g. "[\w@]")
</F>
<pk******@gmail.com> wrote: Oh sorry indentation was messed here...the wordlist = countDict.keys() wordlist.sort() should be outside the word loop.... now def create_words(lines): cnt = 0 spl_set = '[",;<>{}_&?!():-[\.=+*\t\n\r]+' for content in lines: words=content.split() countDict={} wordlist = [] for w in words: w=string.lower(w) if w[-1] in spl_set: w = w[:-1] if w != '': if countDict.has_key(w): countDict[w]=countDict[w]+1 else: countDict[w]=1 wordlist = countDict.keys() wordlist.sort() cnt += 1 if countDict != {}: for word in wordlist: print (word+' '+ str(countDict[word])+'\n')
ok now this is the correct question I am asking...
(a) You might be better off doing:
words = words.lower()
for w in words:
...
instead of calling lower() on each separate word (and note that most
functions from string are deprecated in favour of string methods).
(b) spl_set isn't doing what you might think it is -- it looks like
you've written it as a regexp but your using it as a character set.
What you might want is:
spl_set = '",;<>{}_&?!():-[\.=+*\t\n\r'
and
while w[-1] in spl_set: w = w[:-1]
That loop can be written:
w = w.rstrip(spl_set)
(which by my timings is faster if you have multiple characters from
spl_set at the end of your word, but slower if you have 0 or 1).
--
\S -- si***@chiark.greenend.org.uk -- http://www.chaos.org.uk/~sion/
___ | "Frankly I have no feelings towards penguins one way or the other"
\X/ | -- Arthur C. Clarke
her nu becomeþ se bera eadward ofdun hlæddre heafdes bæce bump bump bump
On 10 Nov 2005 10:43:04 -0800, be************@lycos.com wrote: This can be faster, it avoids doing the same things more times:
from string import maketrans, ascii_lowercase, ascii_uppercase
def create_words(afile): stripper = """'[",;<>{}_&?!():[]\.=+-*\t\n\r^%0123456789/""" mapper = maketrans(stripper + ascii_uppercase, " "*len(stripper) + ascii_lowercase)
good way to prepare for split
countDict = {} for line in afile: for w in line.translate(mapper).split(): if w:
I suspect it's not possible to get '' in the list from somestring.split() if w in countDict: countDict[w] += 1 else: countDict[w] = 1
does that beat the try and get versions? I.e., (untested)
try: countDict[w] += 1
except KeyError: countDict[w] = 1
or
countDict[w] = countDict.get(w, 0) + 1
word_freq = countDict.items() word_freq.sort() for word, freq in word_freq: print word, freq
create_words(file("test.txt"))
If you can load the whole file in memory then it can be made a little faster...
Bear hugs, bearophile
Regards,
Bengt Richter
Bengt Richter enlightened us with: I suspect it's not possible to get '' in the list from somestring.split()
Time to adjust your suspicions: ';abc;'.split(';')
['', 'abc', '']
countDict[w] += 1 else: countDict[w] = 1
does that beat the try and get versions? I.e., (untested) try: countDict[w] += 1 except KeyError: countDict[w] = 1
OPs way is faster. A 'try' block is very fast, but trowing & handling
an exception is slow.
countDict[w] = countDict.get(w, 0) + 1
I think this has the potential of being faster than OPs method,
because it's likely to be fully implemented in C.
Sybren
--
The problem with the world is stupidity. Not saying there should be a
capital punishment for stupidity, but why don't we just take the
safety labels off of everything and let the problem solve itself?
Frank Zappa
Thank you Bengt Richter and Sybren Stuvel for your comments, my little
procedure can be improved a bit in many ways, it was just a first
quickly written version (but it can be enough for a basic usage).
Bengt Richter: good way to prepare for split
Maybe there is a better way, that is putting in just the accepted
letters and accented letters, instead of the not accepted symbols.
I suspect it's not possible to get '' in the list from somestring.split()
You are probably right, the algorithm used is different if you don't
give a splitting string: ' abc a '.split()
['abc', 'a'] '.abc..a.'.split(".")
['', 'abc', '', 'a', '']
does that beat the try and get versions? I.e., (untested)
Yes.
countDict[w] = countDict.get(w, 0) + 1
I think the if-else version is the fastest, I have tested it a long
time ago... You can easely do a speed test to see if I am wrong.
Bear hugs,
bearophile
On Sat, 12 Nov 2005 10:46:53 +0100, Sybren Stuvel <sy*******@YOURthirdtower.com.imagination> wrote: Bengt Richter enlightened us with: I suspect it's not possible to get '' in the list from somestring.split() Time to adjust your suspicions:
';abc;'.split(';')['', 'abc', '']
I know about that one ;-)
I meant somestring.split() just like that -- without a splitter argument.
My suspicion remains ;-) countDict[w] += 1 else: countDict[w] = 1 does that beat the try and get versions? I.e., (untested) try: countDict[w] += 1 except KeyError: countDict[w] = 1
OPs way is faster. A 'try' block is very fast, but trowing & handling an exception is slow.
Yes, so it's data sensitive. Same with the other one not mentioned:
countDict.setdefault(w, [0])[0] += 1 # accum count in length one list vs as int
# too much subscripting work though countDict[w] = countDict.get(w, 0) + 1
I think this has the potential of being faster than OPs method, because it's likely to be fully implemented in C.
Not sure what you mean. Don't forget, all those expressions are evaluated dynamically
step by step with byte codes being switched on in c, but still a bunch of steps. I.e., import dis dis.dis(compile('countDict[w] = countDict.get(w, 0) + 1', '', 'exec'))
1 0 LOAD_NAME 0 (countDict)
3 LOAD_ATTR 1 (get)
6 LOAD_NAME 2 (w)
9 LOAD_CONST 0 (0)
12 CALL_FUNCTION 2
15 LOAD_CONST 1 (1)
18 BINARY_ADD
19 LOAD_NAME 0 (countDict)
22 LOAD_NAME 2 (w)
25 STORE_SUBSCR
26 LOAD_CONST 2 (None)
29 RETURN_VALUE
Looks like if you succeed most of the time without a KeyError, the try/except might be
a tad faster, depending on what the exception block setup/teardown costs.
dis.dis(compile("""\
... try: countDict[w] += 1
... except KeyError: countDict[w] = 1
... """, '', 'exec'))
1 0 SETUP_EXCEPT 20 (to 23)
3 LOAD_NAME 0 (countDict)
6 LOAD_NAME 1 (w)
9 DUP_TOPX 2
12 BINARY_SUBSCR
13 LOAD_CONST 0 (1)
16 INPLACE_ADD
17 ROT_THREE
18 STORE_SUBSCR
19 POP_BLOCK
20 JUMP_FORWARD 29 (to 52)
2 >> 23 DUP_TOP
24 LOAD_NAME 2 (KeyError)
27 COMPARE_OP 10 (exception match)
30 JUMP_IF_FALSE 17 (to 50)
33 POP_TOP
34 POP_TOP
35 POP_TOP
36 POP_TOP
37 LOAD_CONST 0 (1)
40 LOAD_NAME 0 (countDict)
43 LOAD_NAME 1 (w)
46 STORE_SUBSCR
47 JUMP_FORWARD 2 (to 52) 50 POP_TOP
51 END_FINALLY 52 LOAD_CONST 1 (None)
55 RETURN_VALUE
Regards,
Bengt Richter
Bengt Richter enlightened us with: I meant somestring.split() just like that -- without a splitter argument. My suspicion remains ;-)
Mine too ;-)
Sybren
--
The problem with the world is stupidity. Not saying there should be a
capital punishment for stupidity, but why don't we just take the
safety labels off of everything and let the problem solve itself?
Frank Zappa
Fredrik Lundh wrote: Lonnie Princehouse wrote:
"[a-z0-9_]" means "match a single character from the set {a through z, 0 through 9, underscore}".
"\w" should be a bit faster; it's equivalent to "[a-zA-Z0-9_]" (unless you specify otherwise using the locale or unicode flags), but is handled more efficiently by the RE engine.
(you can use \w in sets too, so you can do e.g. "[\w@]")
</F>
The \w does make a small difference, but not as much as I expected.
Surprisingly a standard Python word iterator works just as well, and is
easier to understand than the re version.
Which one is faster depends on the average word length and number of
ignored characters.
Cheers,
Ron
Character count: 100000
Word count: 16477
Average word size: 6.06906597075
word_counter: 0.06820057 (best of 3)
count_words: 0.07333837 (best of 3)
#########
import string
import re
import time
import random
# Create a really ugly n length string to test with.
n = 100000
random.seed(1)
lines = ''.join([ random.choice(string.ascii_letters * 2
+ '_@$&*()#/<>' + ' \n' * 6) for x in range(n) ])
print 'Character count:', n
print 'Word count:', len(lines.split())
print 'Average word size:', float(n)/len(lines.split())
letters = string.lowercase + string.digits + '_@'
def word_iter(text, letters=letters):
ltrs=letters.lower()
wd = ''
for c in text + ' ':
if c in ltrs:
wd += c
elif wd != '':
yield wd
wd = ''
def word_counter(text):
txt = text.lower()
countDict={}
for wd in word_iter(txt):
if wd in countDict:
countDict[wd] += 1
else:
countDict[wd] = 1
return countDict
word_finder = re.compile('[\w@]+', re.I)
def count_words(string, word_finder = word_finder):
# avoid global lookups
countDict = {}
for match in word_finder.finditer(string.lower()):
word = match.group(0)
countDict[word] = countDict.get(word,0) + 1
return countDict
foos = [word_counter, count_words]
r1 = r2 = None
for foo in foos:
best_time = 0
for n in range(3):
t = time.clock()
for line in lines.splitlines():
countDict = foo(line)
tt = time.clock()-t
if tt > best_time: best_time = tt
r1 = r2
r2 = countDict
if r1 != None:
# change to 1 if assert fails to find problem
if 0:
for k in r1.keys():
if r1[k] != r2[k]:
print k,r1[k],r2[k]
assert r1 == r2
print '%s: %.8f (best of %d)' \
% (foo.__name__, best_time, n+1)
Ron Adam wrote: The \w does make a small difference, but not as much as I expected.
that's probably because your benchmark has a lot of dubious overhead:
word_finder = re.compile('[\w@]+', re.I)
no need to force case-insensitive search here; \w looks for both lower-
and uppercase characters.
for match in word_finder.finditer(string.lower()):
since you're using a case-insensitive RE, that lower() call is not necessary.
word = match.group(0)
and findall() is of course faster than finditer() + m.group().
t = time.clock() for line in lines.splitlines(): countDict = foo(line) tt = time.clock()-t
and if you want performance, why are you creating a new dictionary for
each line in the sample?
here's a more optimized RE word finder:
word_finder_2 = re.compile('[\w@]+').findall
def count_words_2(string, word_finder=word_finder_2):
# avoid global lookups
countDict = {}
for word in word_finder(string):
countDict[word] = countDict.get(word,0) + 1
return countDict
with your original test on a slow machine, I get
count_words: 0.29868684 (best of 3)
count_words_2: 0.17244873 (best of 3)
if I call the function once, on the entire sample string, I get
count_words: 0.23096036 (best of 3)
count_words_2: 0.11690620 (best of 3)
</F>
Fredrik Lundh wrote: Ron Adam wrote:
The \w does make a small difference, but not as much as I expected.
that's probably because your benchmark has a lot of dubious overhead:
I think it does what the OP described, but that may not be what he
really needs.
Although the test to find best of n, instead was finding worse of n.
Which explains why I was getting a larger variance than I thought I
should have been getting. word_finder = re.compile('[\w@]+', re.I)
no need to force case-insensitive search here; \w looks for both lower- and uppercase characters.
But the dictionary keys need to be either upper or lower otherwise you
count 'The' separately from 'the'. for match in word_finder.finditer(string.lower()):
since you're using a case-insensitive RE, that lower() call is not necessary.
word = match.group(0)
and findall() is of course faster than finditer() + m.group().
Cool, I don't use re that often so I just used what was posted to test
against. t = time.clock() for line in lines.splitlines(): countDict = foo(line) tt = time.clock()-t
and if you want performance, why are you creating a new dictionary for each line in the sample?
Because that's what the OP apparently wanted. A line by line word
count. I originally did it to get an the over all count and then change
it so it matched the re version that was posted.
here's a more optimized RE word finder:
word_finder_2 = re.compile('[\w@]+').findall
def count_words_2(string, word_finder=word_finder_2): # avoid global lookups countDict = {} for word in word_finder(string): countDict[word] = countDict.get(word,0) + 1 return countDict
with your original test on a slow machine, I get
count_words: 0.29868684 (best of 3) count_words_2: 0.17244873 (best of 3)
if I call the function once, on the entire sample string, I get
count_words: 0.23096036 (best of 3) count_words_2: 0.11690620 (best of 3)
</F>
Wow, a lot bigger difference than on my machine. <curious> An athlon
64 3000+ on winxp. I'm not sure how much difference that would make?
This is what I get after adding the above version to it, with the
lower(). There's not quite as big a difference as you get, but the find
all version is still faster than both the others.
Cheers,
Ron
Character count: 100000
Word count: 16477
Average word size: 6.06906597075
word_counter: 0.06245989 (best of 3)
count_words: 0.07309812 (best of 3)
count_words_2: 0.04981024 (best of 3)
And as count all words...
Character count: 100000
Word count: 16477
Average word size: 6.06906597075
word_counter: 0.05325006 (best of 3)
count_words: 0.05910528 (best of 3)
count_words_2: 0.03748158 (best of 3)
They all improve, but the re find all version is clearly better.
#####################
import string
import re
import time
import random
# Create a really ugly n length string to test with.
# The word length are
n = 100000
random.seed(1)
lines = ''.join([ random.choice(string.ascii_letters * 2
+ '_@$&*()#/<>' + ' \n' * 6) for x in range(n) ])
print 'Character count:', n
print 'Word count:', len(lines.split())
print 'Average word size:', float(n)/len(lines.split())
letters = string.lowercase + string.digits + '_@'
def word_iter(text, letters=letters):
wd = ''
for c in text + ' ':
if c in letters:
wd += c
elif wd != '':
yield wd
wd = ''
def word_counter(text):
countDict={}
for wd in word_iter(text.lower()):
if wd in countDict:
countDict[wd] += 1
else:
countDict[wd] = 1
return countDict
word_finder = re.compile('[\w@]+', re.I).finditer
def count_words(string, word_finder=word_finder):
# avoid global lookups
countDict = {}
for match in word_finder(string.lower()):
word = match.group(0)
countDict[word] = countDict.get(word,0) + 1
return countDict
word_finder_2 = re.compile('[\w@]+').findall
def count_words_2(string, word_finder=word_finder_2):
# avoid global lookups
countDict = {}
for word in word_finder(string.lower()):
countDict[word] = countDict.get(word,0) + 1
return countDict
foos = [word_counter, count_words, count_words_2]
r1 = r2 = None
for foo in foos:
best_time = 1000000 # too large to be useful on purpose
for n in range(3):
t = time.clock()
#for line in lines.splitlines():
countDict = foo(lines)
tt = time.clock()-t
best_time = min(tt, best_time)
r1 = r2
r2 = countDict
if r1 != None:
# change to 1 if assert fails to find problem
if 0:
for k in r1.keys():
if r1[k] != r2[k]:
print k,r1[k],r2[k]
assert r1 == r2
print '%s: %.8f (best of %d)' \
% (foo.__name__, best_time, n+1) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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please let me know. All...
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by: Jay |
last post by:
I have a multi threaded VB.NET application (4 threads) that I use to send
text messages to many, many employees via system.timer at a 5 second
interval. Basically, I look in a SQL table (queue) to...
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by: Joel |
last post by:
(1) Can anyone please tell me the exact meaning of primitive types in
the MSIL context.
(2) Also what is the meaning of the world inline?
(3) What is the meaning of the statement: "You should...
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by: c |
last post by:
Hi every one,
Me and my Cousin were talking about C and C#, I love C and he loves
C#..and were talking C is ...blah blah...C# is Blah Blah ...etc
and then we decided to write a program that...
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by: spider661 |
last post by:
im trying to make a perl file that will take all the info from a spell_us.txt file and place it into an sql file i can import into my database but its not working can anyone help please?
...
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by: ryjfgjl |
last post by:
ExcelToDatabase: batch import excel into database automatically...
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by: isladogs |
last post by:
The next Access Europe meeting will be on Wednesday 6 Mar 2024 starting at 18:00 UK time (6PM UTC) and finishing at about 19:15 (7.15PM).
In this month's session, we are pleased to welcome back...
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by: jfyes |
last post by:
As a hardware engineer, after seeing that CEIWEI recently released a new tool for Modbus RTU Over TCP/UDP filtering and monitoring, I actively went to its official website to take a look. It turned...
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by: PapaRatzi |
last post by:
Hello,
I am teaching myself MS Access forms design and Visual Basic. I've created a table to capture a list of Top 30 singles and forms to capture new entries. The final step is a form (unbound)...
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by: CloudSolutions |
last post by:
Introduction:
For many beginners and individual users, requiring a credit card and email registration may pose a barrier when starting to use cloud servers. However, some cloud server providers now...
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by: Defcon1945 |
last post by:
I'm trying to learn Python using Pycharm but import shutil doesn't work
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by: Shællîpôpï 09 |
last post by:
If u are using a keypad phone, how do u turn on JavaScript, to access features like WhatsApp, Facebook, Instagram....
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by: af34tf |
last post by:
Hi Guys, I have a domain whose name is BytesLimited.com, and I want to sell it. Does anyone know about platforms that allow me to list my domain in auction for free. Thank you
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by: Faith0G |
last post by:
I am starting a new it consulting business and it's been a while since I setup a new website. Is wordpress still the best web based software for hosting a 5 page website? The webpages will be...
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