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This question already has answers here:
Getting SyntaxError for print with keyword argument end=' '
(17 answers)
Closed 8 years ago.
This is the function for printing all values in a nested list (taken from Head first with Python).
def printall(the_list, level):
for x in the_list:
if isinstance(x, list):
printall(x, level=level + 1)
else:
for tab_stop in range(level):
print("\t", end='')
print(x)
The function is working properly.
The function basically prints the values in a list and if there is a nested list then it print it by a tab space.
Just for a better understanding, what does end=' ' do?
I am using Python 3.3.5
For 2.7
f = fi.input( files = 'test2.py', inplace = True, backup = '.bak')
for line in f:
if fi.lineno() == 4:
print line + '\n'
print 'extra line'
else:
print line + '\n'
as of 2.6 fileinput does not support with.
This code appends 3 more lines and prints the appended text on the 3rd new line. and then appends a further 16 empty lines.
The default value of end is \n meaning that after the print statement it will print a new line. So simply stated end is what you want to be printed after the print statement has been executed
Eg: - print ("hello",end=" +") will print hello +
See the documentation for the print function: print()
The content of end is printed after the thing you want to print. By default it contains a newline ("\n") but it can be changed to something else, like an empty string.
I am attempting to create a regular expression pattern for strings similar to the below which are stored in a file. The aim is to get any column for any row, the rows need not be on a single line. So for example, consider the following file:
"column1a","column2a","column
3a,", #entity 1
"column\"this is, a test\"4a"
"column1b","colu
mn2b,","column3b", #entity 2
"column\"this is, a test\"4b"
"column1c,","column2c","column3c", #entity 3
"column\"this is, a test\"4c"
Each entity consists of four columns, column 4 for entity 2 would be "column\"this is, a test\"4b", column 2 for entity 3 would be "column2c". Each column begins with a quote and closes with a quote, however you must be careful because some columns have escaped quotes. Thanks in advance!
You could do like this, ie
Read the whole file.
Split the input according to the newline character which was not preceded by a comma.
Iterate over the spitted elements and again do splitting on the comma (and also the following optional newline character) which was preceded and followed by double quotes.
Code:
import re
with open(file) as f:
fil = f.read()
m = re.split(r'(?<!,)\n', fil.strip())
for i in m:
print(re.split('(?<="),\n?(?=")', i))
Output:
['"column1a"', '"column2a"', '"column3a,"', '"column\\"this is, a test\\"4a"']
['"column1b"', '"column2b,"', '"column3b"', '"column\\"this is, a test\\"4b"']
['"column1c,"', '"column2c"', '"column3c"', '"column\\"this is, a test\\"4c"']
Here is the check..
$ cat f
"column1a","column2a","column3a,",
"column\"this is, a test\"4a"
"column1b","column2b,","column3b",
"column\"this is, a test\"4b"
"column1c,","column2c","column3c",
"column\"this is, a test\"4c"
$ python3 f.py
['"column1a"', '"column2a"', '"column3a,"', '"column\\"this is, a test\\"4a"']
['"column1b"', '"column2b,"', '"column3b"', '"column\\"this is, a test\\"4b"']
['"column1c,"', '"column2c"', '"column3c"', '"column\\"this is, a test\\"4c"']
f is the input file name and f.py is the file-name which contains the python script.
Your problem is terribly familiar to what I have to deal thrice every month :) Except I'm not using python to solve it, but I can 'translate' what I usually do:
text = r'''"column1a","column2a","column
3a,",
"column\"this is, a test\"4a"
"column1a2","column2a2","column3a2","column4a2"
"column1b","colu
mn2b,","column3b",
"column\"this is, a test\"4b"
"column1c,","column2c","column3c",
"column\"this is, a test\"4c"'''
import re
# Number of columns one line is supposed to have
columns = 4
# Temporary variable to hold partial lines
buffer = ""
# Our regex to check for each column
check = re.compile(r'"(?:[^"\\]*|\\.)*"')
# Read the file line by line
for line in text.split("\n"):
# If there's no stored partial line, this is a new line
if buffer == "":
# Check if we get 4 columns and print, if not, put the line
# into buffer so we store a partial line for later
if len(check.findall(line)) == columns:
print matches
else:
# use line.strip() if you need to trim whitespaces
buffer = line
else:
# Update the variable (containing a partial line) with the
# next line and recheck if we get 4 columns
# use line.strip() if you need to trim whitespaces
buffer = buffer + line
# If we indeed get 4, our line is complete and print
# We must not forget to empty buffer now that we got a whole line
if len(check.findall(buffer)) == columns:
print matches
buffer = ""
# Optional; always good to have a safety backdoor though
# If there is a problem with the csv itself like a weird unescaped
# quote, you send it somewhere else
elif len(check.findall(buffer)) > columns:
print "Error: cannot parse line:\n" + buffer
buffer = ""
ideone demo
This question already has answers here:
Print in one line dynamically [duplicate]
(22 answers)
Closed 9 years ago.
The community reviewed whether to reopen this question 4 months ago and left it closed:
Original close reason(s) were not resolved
I was wondering if it was possible to remove items you have printed in Python - not from the Python GUI, but from the command prompt.
e.g.
a = 0
for x in range (0,3):
a = a + 1
b = ("Loading" + "." * a)
print (a)
so it prints
>>>Loading
>>>Loading.
>>>Loading..
>>>Loading...
But, my problem is I want this all on one line, and for it it remove it self when something else comes along. So instead of printing "Loading", "Loading.", "Loading... I want it to print "Loading.", then it removes what is on the line and replaces it with "Loading.." and then removes "Loading.." and replaces it (on the same line) with "Loading...". It's kind of hard to describe.
p.s I have tried to use the Backspace character but it doesn't seem to work ("\b")
Just use CR to go to beginning of the line.
import time
for x in range (0,5):
b = "Loading" + "." * x
print (b, end="\r")
time.sleep(1)
One way is to use ANSI escape sequences:
import sys
import time
for i in range(10):
print("Loading" + "." * i)
sys.stdout.write("\033[F") # Cursor up one line
time.sleep(1)
Also sometimes useful (for example if you print something shorter than before):
sys.stdout.write("\033[K") # Clear to the end of line
import sys
import time
a = 0
for x in range (0,3):
a = a + 1
b = ("Loading" + "." * a)
# \r prints a carriage return first, so `b` is printed on top of the previous line.
sys.stdout.write('\r'+b)
time.sleep(0.5)
print (a)
Note that you might have to run sys.stdout.flush() right after sys.stdout.write('\r'+b) depending on which console you are doing the printing to have the results printed when requested without any buffering.
I have done this operation millions of times, just using the + operator! I have no idea why it is not working this time, it is overwriting the first part of the string with the new one! I have a list of strings and just want to concatenate them in one single string! If I run the program from Eclipse it works, from the command-line it doesn't!
The list is:
["UNH+1+XYZ:08:2:1A+%CONVID%'&\r", "ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&\r", "DUM'&\r"]
I want to discard the first and the last elements, the code is:
ediMsg = ""
count = 1
print "extract_the_info, lineList ",lineList
print "extract_the_info, len(lineList) ",len(lineList)
while (count < (len(lineList)-1)):
temp = ""
# ediMsg = ediMsg+str(lineList[count])
# print "Count "+str(count)+" ediMsg ",ediMsg
print "line value : ",lineList[count]
temp = lineList[count]
ediMsg += " "+temp
print "ediMsg : ",ediMsg
count += 1
print "count ",count
Look at the output:
extract_the_info, lineList ["UNH+1+XYZ:08:2:1A+%CONVID%'&\r", "ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&\r", "DUM'&\r"]
extract_the_info, len(lineList) 8
line value : ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&
ediMsg : ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&
count 2
line value : DUM'&
DUM'& : ORG+1A+77499505:ABC+++A+FR:EUR++123+1A'&
count 3
Why is it doing so!?
While the two answers are correct (use " ".join()), your problem (besides very ugly python code) is this:
Your strings end in "\r", which is a carriage return. Everything is fine, but when you print to the console, "\r" will make printing continue from the start of the same line, hence overwrite what was written on that line so far.
You should use the following and forget about this nightmare:
''.join(list_of_strings)
The problem is not with the concatenation of the strings (although that could use some cleaning up), but in your printing. The \r in your string has a special meaning and will overwrite previously printed strings.
Use repr(), as such:
...
print "line value : ", repr(lineList[count])
temp = lineList[count]
ediMsg += " "+temp
print "ediMsg : ", repr(ediMsg)
...
to print out your result, that will make sure any special characters doesn't mess up the output.
'\r' is the carriage return character. When you're printing out a string, a '\r' will cause the next characters to go at the start of the line.
Change this:
print "ediMsg : ",ediMsg
to:
print "ediMsg : ",repr(ediMsg)
and you will see the embedded \r values.
And while your code works, please change it to the one-liner:
ediMsg = ' '.join(lineList[1:-1])
Your problem is printing, and it is not string manipulation. Try using '\n' as last char instead of '\r' in each string in:
lineList = [
"UNH+1+TCCARQ:08:2:1A+%CONVID%'&\r",
"ORG+1A+77499505:PARAF0103+++A+FR:EUR++11730788+1A'&\r",
"DUM'&\r",
"FPT+CC::::::::N'&\r",
"CCD+CA:5132839000000027:0450'&\r",
"CPY+++AF'&\r",
"MON+712:1.00:EUR'&\r",
"UNT+8+1'\r"
]
I just gave it a quick look. It seems your problem arises when you are printing the text. I haven't done such things for a long time, but probably you only get the last line when you print. If you check the actual variable, I'm sure you'll find that the value is correct.
By last line, I'm talking about the \r you got in the text strings.
I have a text file which a lot of random occurrences of the string #STRING_A, and I would be interested in writing a short script which removes only some of them. Particularly one that scans the file and once it finds a line which starts with this string like
#STRING_A
then checks if 3 lines backwards there is another occurrence of a line starting with the same string, like
#STRING_A
#STRING_A
and if it happens, to delete the occurrence 3 lines backward. I was thinking about bash, but I do not know how to "go backwards" with it. So I am sure that this is not possible with bash. I also thought about python, but then I should store all information in memory in order to go backwards and then, for long files it would be unfeasible.
What do you think? Is it possible to do it in bash or python?
Thanks
Funny that after all these hours nobody's yet given a solution to the problem as actually phrased (as #John Machin points out in a comment) -- remove just the leading marker (if followed by another such marker 3 lines down), not the whole line containing it. It's not hard, of course -- here's a tiny mod as needed of #truppo's fun solution, for example:
from itertools import izip, chain
f = "foo.txt"
for third, line in izip(chain(" ", open(f)), open(f)):
if third.startswith("#STRING_A") and line.startswith("#STRING_A"):
line = line[len("#STRING_A"):]
print line,
Of course, in real life, one would use an iterator.tee instead of reading the file twice, have this code in a function, not repeat the marker constant endlessly, &c;-).
Of course Python will work as well. Simply store the last three lines in an array and check if the first element in the array is the same as the value you are currently reading. Then delete the value and print out the current array. You would then move over your elements to make room for the new value and repeat. Of course when the array is filled you'd have to make sure to continue to move values out of the array and put in the newly read values, stopping to check each time to see if the first value in the array matches the value you are currently reading.
Here is a more fun solution, using two iterators with a three element offset :)
from itertools import izip, chain, tee
f1, f2 = tee(open("foo.txt"))
for third, line in izip(chain(" ", f1), f2):
if not (third.startswith("#STRING_A") and line.startswith("#STRING_A")):
print line,
Why shouldn't it possible in bash? You don't need to keep the whole file in memory, just the last three lines (if I understood correctly), and write what's appropriate to standard-out. Redirect that into a temporary file, check that everything worked as expected, and overwrite the source file with the temporary one.
Same goes for Python.
I'd provide a script of my own, but that wouldn't be tested. ;-)
As AlbertoPL said, store lines in a fifo for later use--don't "go backwards". For this I would definitely use python over bash+sed/awk/whatever.
I took a few moments to code this snippet up:
from collections import deque
line_fifo = deque()
for line in open("test"):
line_fifo.append(line)
if len(line_fifo) == 4:
# "look 3 lines backward"
if line_fifo[0] == line_fifo[-1] == "#STRING_A\n":
# get rid of that match
line_fifo.popleft()
else:
# print out the top of the fifo
print line_fifo.popleft(),
# don't forget to print out the fifo when the file ends
for line in line_fifo: print line,
This code will scan through the file, and remove lines starting with the marker. It only keeps only three lines in memory by default:
from collections import deque
def delete(fp, marker, gap=3):
"""Delete lines from *fp* if they with *marker* and are followed
by another line starting with *marker* *gap* lines after.
"""
buf = deque()
for line in fp:
if len(buf) < gap:
buf.append(line)
else:
old = buf.popleft()
if not (line.startswith(marker) and old.startswith(marker)):
yield old
buf.append(line)
for line in buf:
yield line
I've tested it with:
>>> from StringIO import StringIO
>>> fp = StringIO('''a
... b
... xxx 1
... c
... xxx 2
... d
... e
... xxx 3
... f
... g
... h
... xxx 4
... i''')
>>> print ''.join(delete(fp, 'xxx'))
a
b
xxx 1
c
d
e
xxx 3
f
g
h
xxx 4
i
This "answer" is for lyrae ... I'll amend my previous comment: if the needle is in the first 3 lines of the file, your script will either cause an IndexError or access a line that it shouldn't be accessing, sometimes with interesting side-effects.
Example of your script causing IndexError:
>>> lines = "#string line 0\nblah blah\n".splitlines(True)
>>> needle = "#string "
>>> for i,line in enumerate(lines):
... if line.startswith(needle) and lines[i-3].startswith(needle):
... lines[i-3] = lines[i-3].replace(needle, "")
...
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
IndexError: list index out of range
and this example shows not only that the Earth is round but also why your "fix" to the "don't delete the whole line" problem should have used .replace(needle, "", 1) or [len(needle):] instead of .replace(needle, "")
>>> lines = "NEEDLE x NEEDLE y\nnoddle\nnuddle\n".splitlines(True)
>>> needle = "NEEDLE"
>>> # Expected result: no change to the file
... for i,line in enumerate(lines):
... if line.startswith(needle) and lines[i-3].startswith(needle):
... lines[i-3] = lines[i-3].replace(needle, "")
...
>>> print ''.join(lines)
x y <<<=== whoops!
noddle
nuddle
<<<=== still got unwanted newline in here
>>>
My awk-fu has never been that good... but the following may provide you what you're looking for in a bash-shell/shell-utility form:
sed `awk 'BEGIN{ORS=";"}
/#STRING_A/ {
if(LAST!="" && LAST+3 >= NR) print LAST "d"
LAST = NR
}' test_file` test_file
Basically... awk is producing a command for sed to strip certain lines. I'm sure there's a relatively easy way to make awk do all of the processing, but this does seem to work.
The bad part? It does read the test_file twice.
The good part? It is a bash/shell-utility implementation.
Edit: Alex Martelli points out that the sample file above might have confused me. (my above code deletes the whole line, rather than the #STRING_A flag only)
This is easily remedied by adjusting the command to sed:
sed `awk 'BEGIN{ORS=";"}
/#STRING_A/ {
if(LAST!="" && LAST+3 >= NR) print LAST "s/#STRING_A//"
LAST = NR
}' test_file` test_file
This may be what you're looking for?
lines = open('sample.txt').readlines()
needle = "#string "
for i,line in enumerate(lines):
if line.startswith(needle) and lines[i-3].startswith(needle):
lines[i-3] = lines[i-3].replace(needle, "")
print ''.join(lines)
this outputs:
string 0 extra text
string 1 extra text
string 2 extra text
string 3 extra text
--replaced -- 4 extra text
string 5 extra text
string 6 extra text
#string 7 extra text
string 8 extra text
string 9 extra text
string 10 extra text
In bash you can use sort -r filename and tail -n filename to read the file backwards.
$LINES=`tail -n filename | sort -r`
# now iterate through the lines and do your checking
I would consider using sed. gnu sed supports definition of line ranges. if sed would fail, then there is another beast - awk and I'm sure you can do it with awk.
O.K. I feel I should put my awk POC. I could not figure out to use sed addresses. I have not tried combination of awk+sed, but it seems to me it's overkill.
my awk script works as follows:
It reads lines and stores them into 3 line buffer
once desired pattern is found (/^data.*/ in my case), the 3-line buffer is looked up to check, whether desired pattern has been seen three lines ago
if pattern has been seen, then 3 lines are scratched
to be honest, I would probably go with python also, given that awk is really awkward.
the AWK code follows:
function max(a, b)
{
if (a > b)
return a;
else
return b;
}
BEGIN {
w = 0; #write index
r = 0; #read index
buf[0, 1, 2]; #buffer
}
END {
# flush buffer
# start at read index and print out up to w index
for (k = r % 3; k r - max(r - 3, 0); k--) {
#search in 3 line history buf
if (match(buf[k % 3], /^data.*/) != 0) {
# found -> remove lines from history
# by rewriting them -> adjust write index
w -= max(r, 3);
}
}
buf[w % 3] = $0;
w++;
}
/^.*/ {
# store line into buffer, if the history
# is full, print out the oldest one.
if (w > 2) {
print buf[r % 3];
r++;
buf[w % 3] = $0;
}
else {
buf[w] = $0;
}
w++;
}