I want to create a string using an integer appended to it, in a for loop. Like this:

for i in range(1, 11): string = "string" + i

But it returns an error:

TypeError: unsupported operand type(s) for +: 'int' and 'str'

What's the best way to concatenate the string and integer?

9 Answers

NOTE:

The method used in this answer (backticks) is deprecated in later versions of Python 2, and removed in Python 3. Use the str() function instead.

You can use:

string = 'string' for i in range(11): string +=`i` print string

It will print string012345678910.

To get string0, string1 ..... string10 you can use this as YOU suggested:

>>> string = "string" >>> [string+`i` for i in range(11)]

For Python 3

You can use:

string = 'string' for i in range(11): string += str(i) print string

It will print string012345678910.

To get string0, string1 ..... string10, you can use this as YOU suggested:

>>> string = "string" >>> [string+str(i) for i in range(11)]

for i in range (1,10): string="string"+str(i)

To get string0, string1 ..... string10, you could do like

>>> ["string"+str(i) for i in range(11)] ['string0', 'string1', 'string2', 'string3', 'string4', 'string5', 'string6', 'string7', 'string8', 'string9', 'string10']

for i in range[1,10]: string = "string" + str(i)

The str(i) function converts the integer into a string.

string = 'string%d' % (i,)

for i in range(11): string = "string{0}".format(i)

You did (range[1,10]):

a TypeError since brackets denote an index (a[3]) or a slice (a[3:5]) of a list,
a SyntaxError since [1,10] is invalid, and
a double off-by-one error since range(1,10) is [1, 2, 3, 4, 5, 6, 7, 8, 9], and you seem to want [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

And string = "string" + i is a TypeError since you can't add an integer to a string (unlike JavaScript).

Look at the documentation for Python's new string formatting method. It is very powerful.

You can use a generator to do this!

def sequence_generator(limit): """ A generator to create strings of pattern -> string1,string2..stringN """ inc = 0 while inc < limit: yield 'string' + str(inc) inc += 1 # To generate a generator. Notice I have used () instead of [] a_generator = (s for s in sequence_generator(10)) # To generate a list a_list = [s for s in sequence_generator(10)] # To generate a string a_string = '['+ ", ".join(s for s in sequence_generator(10)) + ']'

If we want output like 'string0123456789' then we can use the map function and join method of string.

>>> 'string' + "".join(map(str, xrange(10))) 'string0123456789'

If we want a list of string values then use the list comprehension method.

>>> ['string'+i for i in map(str,xrange(10))] ['string0', 'string1', 'string2', 'string3', 'string4', 'string5', 'string6', 'string7', 'string8', 'string9']

Note:

Use xrange() for Python 2.x.

Use range() for Python 3.x.

I did something else.

I wanted to replace a word, in lists of lists, that contained phrases.

I wanted to replace that string / word with a new word that will be a join between string and number, and that number / digit will indicate the position of the phrase / sublist / lists of lists.

That is, I replaced a string with a string and an incremental number that follow it.

myoldlist_1 = [[' myoldword'], [''], ['tttt myoldword'], ['jjjj ddmyoldwordd']] No_ofposition = [] mynewlist_2 = [] for i in xrange(0, 4, 1): mynewlist_2.append([x.replace('myoldword', "%s" % i + "_mynewword") for x in myoldlist_1[i]]) if len(mynewlist_2[i]) > 0: No_ofposition.append(i) mynewlist_2 No_ofposition

Concatenation of a string and integer is simple: just use