this is my first message here, I hope I will not commit any mistake.
I am writing a python 2.7 script which performs comparisons between lines from a long list of lines provided as an external input file. Some of these lines contain just numbers, and on those I perform simple sums after their retrieval via getline.linecache.
My problem is that after a certain number of lines I am getting the error:
ValueError: invalid literal for int() with base 10
I do understand that somehow this has to do with the fact that there is some problem when I try to convert the lines retrieved to the integer type, but according to what I read each line should be retrieved from a memory database as a string, and indeed if I try to print the type of the values retrieved I get str. I printed the problematic values in order to understand why they failed to be converted to int: at first i included some semantic mistakes (I was taking some wrong lines, which were containing letters, and this of course failed to be converted to int), but still I get the error on merely numerical strings. On all of those numerical strings, I tried len(linecache.getline('input', line_n)) to see if any extra characters were present, but I just found '\n', which does not give any problems when converting from str to int.
My input file is made by a series of lines, some numerical some not; here are few lines:
1
id3021-a
1
129485768
129485769
2
id2034
102
944709842
944709848
For examples, line 4 here can be retrieved, but not converted to int. How could I convert str to int without getting errors?
I found the solution! Adding a '0' to the beginning of the string fixes the problem (I do not know why, the problematic lines were not empty):
int('0' + linecache.getline('input', line_n))
See here: Trouble converting string to int in Django/Python
Related
I am trying to convert a string number from one MySQL table to another.
I have used the following on many occasions, however it does not seem to be working in this instance and I am unsure as to why. The string it is converting is 50,000.00.
With that formula in tMap it produces the following error:
When I look at the code on 3031:
So something is just not functioning as I expect. Any help would be great.
In regex syntax, "$" indicates the end of the string. In your regex, you are trying to remove any character after the end of the string which is not a number or a dot, which won't work, so the "," is never removed from your string, resulting in a conversion error.
You can do this:
new BigDecimal(row1.Trade_Amount.replaceAll("[^\\d.]", ""))
I am learning python. while doing an exercise, i encountered a problem
Python TypeError: not all arguments converted during string formatting
I tried to refer and change the code but nothing worked well.
my_list=input(list)
for num in my_list:
if(num%2==0):
print(f"{num} is even number")
else:
print(f'{num} is odd number')
I expected the output should be either num is even number or num is odd number for every number in the list. Unfortunately I get the error message above.
How can I fix this?
my_list=map(int, input().split())
for num in my_list:
if(num%2==0):
print(f"{num} is even number")
else:
print(f'{num} is odd number')
give the input space seperated integers for eg : "1 2 3 4" without quotes. The output should be as expected.
The error is probably due to the input format which you might be giving in list for mat i.e "[1,2,3,4]" which not implicitly decoded by string formatting in python. To input list of numbers you might want to use the way i have mentioned in the code.
i am facing issue while converting unicode data into national characters.
When i convert the Unicode data into national using national-of function, some junk character like # is appended after the string.
E.g
Ws-unicode pic X(200)
Ws-national pic N(600)
--let the value in Ws-Unicode is これらの変更は. getting from java end.
move function national-of ( Ws-unicode ,1208 ) to Ws-national.
--after converting value is like これらの変更は #.
i do not want the extra # character added after conversion.
please help me to find out the possible solution, i have tried to replace N'#' with space using inspect clause.
it worked well but failed in some specific scenario like if we have # in input from user end. in that case genuine # also converted to space.
Below is a snippet of code I used to convert EBCDIC to UTF. Before I was capturing string lengths, I was also getting # symbols:
STRING
FUNCTION DISPLAY-OF (
FUNCTION NATIONAL-OF (
WS-EBCDIC-STRING(1:WS-XML-EBCDIC-LENGTH)
WS-EBCDIC-CCSID
)
WS-UTF8-CCSID
)
DELIMITED BY SIZE
INTO WS-UTF8-STRING
WITH POINTER WS-XML-UTF8-LENGTH
END-STRING
SUBTRACT 1 FROM WS-XML-UTF8-LENGTH
What this code does is string the UTF8 representation of the EBCIDIC string into another variable. The WITH POINTER clause will capture the new length of the string + 1 (+ 1 because the pointer is positioned to the next position after the string ended).
Using this method, you should be able to know exactly how long second string is and use that string with the exact length.
That should remove the unwanted #s.
EDIT:
One thing I forgot to mention, in my case, the # signs were actually EBCDIC low values when viewing the actual hex on the mainframe
Use inspect with reverse and stop after first occurence of #
So I'm reading multiple text files in Matlab that have, in their first columns, a column of "times". These times are either in the format 'MM:SS.milliseconds' (sorry if that's not the proper way to express it) where for example the string '29:59.9' would be (29*60)+(59)+(.9) = 1799.9 seconds, or in the format of straight seconds.milliseconds, where '29.9' would mean 29.9 seconds. The format is the same for a single file, but varies across different files. Since I would like the times to be in the second format, I would like to check if the format of the strings match the first format. If it doesn't match, then convert it, otherwise, continue. The code below is my code to convert, so my question is how do I approach checking the format of the string? In otherwords, I need some condition for an if statement to check if the format is wrong.
%% Modify the textdata to convert time to seconds
timearray = textdata(2:end, 1);
if (timearray(1, 1) %{has format 'MM.SS.millisecond}%)
datev = datevec(timearray);
newtime = (datev(:, 5)*60) + (datev(:, 6));
elseif(timearray(1, 1) %{has format 'SS.millisecond}%)
newtime = timearray;
You can use regular expressions to help you out. Regular expressions are methods of specifying how to search for particular patterns in strings. As such, you want to find if a string follows the formats of either:
xx:xx.x
or:
xx.x
The regular expression syntax for each of these is defined as the following:
^[0-9]+:[0-9]+\.[0-9]+
^[0-9]+\.[0-9]+
Let's step through how each of these work.
For the first one, the ^[0-9]+ means that the string should start with any number (^[0-9]) and the + means that there should be at least one number. As such, 1, 2, ... 10, ... 20, ... etc. is valid syntax for this beginning. After the number should be separated by a :, followed by another sequence of numbers of at least one or more. After, there is a . that separates them, then this is followed by another sequence of numbers. Notice how I used \. to specify the . character. Using . by itself means that the character is a wildcard. This is obviously not what you want, so if you want to specify the actual . character, you need to prepend a \ to the ..
For the second one, it's almost the same as the first one. However, there is no : delimiter, and we only have the . to work with.
To invoke regular expressions, use the regexp command in MATLAB. It is done using:
ind = regexp(str, expression);
str represents the string you want to check, and expression is a regular expression that we talked about above. You need to make sure you encapsulate your expression using single quotes. The regular expression is taken in as a string. ind would this return the starting index of your string of where the match was found. As such, when we search for a particular format, ind should either be 1 indicating that we found this search at the beginning of the string, or it returns empty ([]) if it didn't find a match. Here's a reproducible example for you:
B = {'29:59.9', '29.9', '45:56.8', '24.5'};
for k = 1 : numel(B)
if (regexp(B{k}, '^[0-9]+:[0-9]+\.[0-9]+') == 1)
disp('I''m the first case!');
elseif (regexp(B{k}, '^[0-9]+\.[0-9]+') == 1)
disp('I''m the second case!');
end
end
As such, the code should print out I'm the first case! if it follows the format of the first case, and it should print I'm the second case! if it follows the format of the second case. As such, by running this code, we get:
I'm the first case!
I'm the second case!
I'm the first case!
I'm the second case!
Without knowing how your strings are formatted, I can't do the rest of it for you, but this should be a good start for you.
I have some code where I am converting some data elements in a flat file. I save the old:new values to a hash which is written to a file at the end of processing. On subsequence execution, I reload into a hash so I can reuse previously converted values on additional data files. I also save the last conversion value so if I encounter an unconverted value, I can assign it a new converted value and add it to the hash.
I had used this code before (back in Feb) on six files with no issues. I have a variable that is set to ZCKL0 (last character is a zero) which is retrieved from a file holding the last used value. I apply the increment operator
...
$data{$olddata} = ++$dataseed;
...
and the resultant value in $dataseed is 1 instead of ZCKL1. The original starting seed value was ZAAA0.
What am I missing here?
Do you use the $dataseed variable in a numeric context in your code?
From perlop:
If you increment a variable that is
numeric, or that has ever been used in
a numeric context, you get a normal
increment. If, however, the variable
has been used in only string contexts
since it was set, and has a value that
is not the empty string and matches
the pattern /^[a-zA-Z][0-9]\z/ , the
increment is done as a string,
preserving each character within its
range.
As prevously mentioned, ++ on strings is "magic" in that it operates differently based on the content of the string and the context in which the string is used.
To illustrate the problem and assuming:
my $s='ZCL0';
then
print ++$s;
will print:
ZCL1
while
$s+=0; print ++$s;
prints
1
NB: In other popular programming languages, the ++ is legal for numeric values only.
Using non-intuitive, "magic" features of Perl is discouraged as they lead to confusing and possibly unsupportable code.
You can write this almost as succinctly without relying on the magic ++ behavior:
s/(\d+)$/ $1 + 1 /e
The e flag makes it an expression substitution.