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.
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.]", ""))
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.
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
I'v read and re-read the help about the function sprintf in matlab but I do not understand everything about this function and the format they talk about.
I was asking myself the logic behind the function formats.
If I run the example
sprintf('%05d%s%02d%s%02d',546,'.',1,'.',3)
I get
00546.01.03
which is logic, since the first number (546) is written as an integer and with 5 digits, the second is a character, and so on... But if now I try this
sprintf('%05d%s%02d%s%02d',546,'.',1,'.',3,4)
I get
00546.01.0300004
the first part is the same as above... But the last part of it (00004) has the format '%05d', that corresponds to the first format I entered in the function's arguments. My question is then Does the first format become the 'default' format ?
By trying this
sprintf('%05d%s%02d%s%02d',546,'.',1,'.',3,4,56)
and getting this
00546.01.03000048
I think the answer is no... But why ? And what is then the logic behind those arguments?
Thanks for your help !
You are providing sprintf more arguments than there are %s in the format string. Therefore, sprintf re-uses the format string from begining:
sprintf('%05d%s%02d%s%02d',546,'.',1,'.',3,4,56)
result:
00546.01.03000048
^
starting fromat anew printing 00004 for %05d with 4
The final '8' character is 56 printed as '%s' (if you want to check it out the ascii code of '8' (the char) is 56!)
I want to convert int to string and then concatenate dot with it. Here is the formula
totext({#SrNo})+ "."
It works perfectly but not what i want. I want to show at as
1.
but it shows me in this way
1.00.
it means that when i try to convert int to string it convert it into number with precision of two decimal zeros. Can someone tell me how can i show it in proper format. For information i want to tell you that SrNo is running total.
ToText(x, y, z, w) Function can use
x=The number to convert to text
y=The number of decimal places to include in result (optional). The value will be rounded to that decimal place.
z=The character to use as the thousands separator. If you don’t specify one, it will use your application default. (Optional.)
w=The character to use as the decimal separator. If you don’t specify one, it will use your application default. (Optional.)
Examples
ToText(12345.678) = > “12345.678″
ToText(12345.678,2) = > “12345.67″
ToText(12345.678,0) = > “12345″
You can try this :
totext({fieldname},0)
Ohhh I got the answer it was so simple.
totext takes 4 parameters
First parameter is value which is going to be converted
Second parameter is number of decimal previsions.
Third parameter is decimal separator. like (1,432.123) here dot(.) is third parameter.
Forth parameter is thousand separator. like (1,432) here comma(,) is forth parameter.
Example{
totext("1,432.1234",2) results 1,432.12
totext("1,432.1234",2,' " ') results 1,432"1234
totext("1,432.1234",2,' " ', ' : ') results 1:432,1234
}
Although i think this example may be not so good but i just want to give you an idea. This is for int conversion for date it has 2 parameters.
value to be converted and format of date.