str.format() raises KeyError

Question

The following code raises a KeyError exception:

addr_list_formatted = []
addr_list_idx = 0

for addr in addr_list: # addr_list is a list
    addr_list_idx = addr_list_idx + 1
    addr_list_formatted.append("""
        "{0}"
        {
        "gamedir"  "str"
        "address"  "{1}"
        }
    """.format(addr_list_idx, addr))

Why?

I am using Python 3.1.

You could probably accomplish the same task with a list comprehension (use `enumerate` to get the indeces). — SwiftsNamesake, May 12 '15 at 16:35
**See also (duplicate)** https://stackoverflow.com/questions/35574349/python-format-string-with-custom-delimiters — dreftymac, Jul 01 '18 at 14:55

score 155 · Accepted Answer · edited Mar 30 '18 at 16:41

155

The problem is those { and } characters you have there that don't specify a key for formatting. You need to double them up, so change your code to:

addr_list_formatted.append("""
    "{0}"
    {{
    "gamedir"  "str"
    "address"  "{1}"
    }}
""".format(addr_list_idx, addr))

edited Mar 30 '18 at 16:41

MountainDrew

383
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18

answered May 02 '10 at 22:30

Lasse V. Karlsen

366,661
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12

What if someone wanted to use JSON in Python? – fijiaaron Jun 24 '16 at 00:47
4

@fijiaaron the double `{` in the answer is just to tell the `format` method that there is no key to format here (so they are escaped in the formated string and it shouldn't be a problem for build a JSON that way). Alternatively there is other efficient ways to manipulate strings, like the `join` method : `"".join(['{"', var_name, '":', value, '}'])` – mgc Jun 24 '16 at 00:57
I almost lost my mind until figured whats the problem – Dmitry Kankalovich Jul 16 '20 at 01:53
2

@DmitryKankalovich You have lost your mind, Stack Overflow, and everybody here, is just a figment of your imagination. (that's what I keep telling myself at least) :) – Lasse V. Karlsen Jul 16 '20 at 11:11

str.format() raises KeyError

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