Reputation: 583

Casting an int to a string in Python

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

Upvotes: 53

Answers (6)

Ferdinand Beyer

Reputation: 67177

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).

Upvotes: 19

user5994461

Reputation: 7128

Works on Python 3 and Python 2.7

for i in range(50):
    filename = "example{:03d}.txt".format(i)
    print(filename)

Use format {:03d} to have 3 digits with leading 0, it's a great trick to have the files appear in the right order when looking at them in the explorer. Leave brackets empty {} for default formatting.

# output
example000.txt
example001.txt
example002.txt
example003.txt

Upvotes: 0

Florian Mayer

Reputation: 3081

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

Please see the Python documentation: https://docs.python.org/3/library/functions.html#str

Upvotes: 97

Tony Veijalainen

Reputation: 5555

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()

Upvotes: 0