Python create string with number

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?

asked May 17, 2010 at 7:52

6

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]]

answered Aug 21, 2013 at 17:45

8

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']

answered May 17, 2010 at 7:53

YOUYOU

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5

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

The str[i] function converts the integer into a string.

answered May 17, 2010 at 7:53

Rizwan KassimRizwan Kassim

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0

string = 'string%d' % [i,]

answered May 17, 2010 at 7:53

2

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.

answered May 17, 2010 at 8:17

Tim PietzckerTim Pietzcker

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1

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]] + ']'

answered Jul 3, 2015 at 3:01

0

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.

answered Jan 5, 2015 at 9:10

Vivek SableVivek Sable

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1

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

answered Apr 29, 2016 at 14:20

Concatenation of a string and integer is simple: just use

abhishek+str[2]

answered Apr 1, 2016 at 9:34

HiroHiro

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1

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