Casting an int to a string in Python

PythonStringIntegerConcatenation

Python Problem Overview


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

Python Solutions


Solution 1 - Python

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

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

Solution 2 - Python

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).

Solution 3 - Python

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.

Solution 4 - Python

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

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

Solution 5 - Python

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()

Solution 6 - Python

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

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionpythonTAView Question on Stackoverflow
Solution 1 - PythonFlorian MayerView Answer on Stackoverflow
Solution 2 - PythonFerdinand BeyerView Answer on Stackoverflow
Solution 3 - PythonFrédéric HamidiView Answer on Stackoverflow
Solution 4 - PythonnmichaelsView Answer on Stackoverflow
Solution 5 - PythonTony VeijalainenView Answer on Stackoverflow
Solution 6 - Pythonuser5994461View Answer on Stackoverflow