Casting an int to a string in Python

问题

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

回答1:

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

Please see the Python documentation: https://docs.python.org/2/library/functions.html#str

回答2:

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).

回答3:

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.

回答4:

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

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

回答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()

来源：https://stackoverflow.com/questions/3944876/casting-an-int-to-a-string-in-python