Casting an int to a string in Python

Question

I want to be able to generate a number of text files with the names fileX.txt where X is some integer:

for i in range(key):
    filename = "ME" + i + ".txt" //Error here! Can't concat a string and int
    filenum = filename
    filenum = open(filename , 'w')  

Does anyone else know how to do the filename = "ME" + i part so I get a list of files with the names: "ME0.txt" , "ME1.txt" , "ME2.txt" , and etc

Answer 1

For Python versions prior to 2.6, use the string formatting operator %:

filename = "ME%d.txt" % i

For 2.6 and later, use the str.format() method:

filename = "ME{0}.txt".format(i)

Though the first example still works in 2.6, the second one is preferred.

If you have more than 10 files to name this way, you might want to add leading zeros so that the files are ordered correctly in directory listings:

filename = "ME%02d.txt" % i
filename = "ME{0:02d}.txt".format(i)

This will produce file names like ME00.txt to ME99.txt. For more digits, replace the 2 in the examples with a higher number (eg, ME{0:03d}.txt).

Answer 2

Either:

"ME" + str(i)

Or:

"ME%d" % i

The second one is usually preferred, especially if you want to build a string from several tokens.

Answer 3

x = 1
y = "foo" + str(x)

Please read the Python documentation: http://docs.python.org/

Answer 4

You can use str() to cast it, or formatters:

"ME%d.txt" % (num,)

Answer 5

Here answer for your code as whole:

key =10

files = ("ME%i.txt" % i for i in range(key))

#opening
files = [ open(filename, 'w') for filename in files]

# processing
for i, file in zip(range(key),files):
    file.write(str(i))
# closing
for openfile in files:
    openfile.close()