Get the iterator value while using a for structure in Python

Question

I must find a way to get the item position when iterating over a list using a for structure

other_list = ["line1", "line2", "line3", ... , "line125k+"]
#contains 125k+ items from a txtFile.readlines()

list = ["item1", "item2", "item3"]
#contains 35 items

dict = {"key1":["value1"], "key2":["value2"], "key3":["value3"]}
#contains 35 items too

For each value inside my dict, i got a key that have a correspondent item in the list.

list = ["10", "20", "30"]`
dict = {"19":["value1"], "29":["value2"], "39":["value3"]}

the dict's first key, "19", corresponds to the "10" inside the other list..

Example:
dict[0] corresponds to list[0]
dict[1] corresponds to list[1]
... and so on.

So i have to get the item position while using a for structure, so then i can access the correspondent key in the dict, and use the dict's value in a replace()

#replace tax1 value
for item in list:
    pos_item = item.getPosition() # pos_item = getIteratorValue()
    #how can i assign the dict value to a variable?
    dict_value = dict[pos_item][value]
    #use one variable to search and the other as a replacement
    other_list[pos_item].replace("0,00", dict_value)

I think it's not possible to get the index in a foreach loop because of the iterator model. The only thing you could do is set a variable to 0 before the loop and increment it every single run... — Luca Schimweg
– Luca Schimweg, Commented Jul 18, 2015 at 20:22
Since dictionaries are not ordered you can not compare its item's position with a list elements! — Kasravnd
– Kasravnd, Commented Jul 18, 2015 at 20:23
Based on what you want to do you have 2 choice, at first use a OrderdDict instead of a dict or sort your dict items and work with the sorted result which is not a dictionary anymore and is a list! — Kasravnd
– Kasravnd, Commented Jul 18, 2015 at 20:28
You could use enumerate(), to find out the current position in the iteration . Obviously, using index won't allow to get dict's items — jfs
– jfs, Commented Jul 18, 2015 at 20:33

kirbyfan64sos · Accepted Answer · 2015-07-18 21:28:43Z

1

There are three issues here, not one. In Python (and most other languages), dictionaries aren't sorted, and replace returns a new string. They have no order. In order to get around this, you can use an OrderedDict and do something like:

# The dictionary.
dct = OrderedDict([('key1', 'value1'), ('key2', 'value2')]
for pos_item, (item, dict_values) in enumerate(zip(lst, dct.values())):
   dict_value = dict_values[value]
   other_list[pos_item] = other_list[pos_item].replace('0,00', dict_value)

Look into enumerate and zip.

Also notice that I renamed dict to dct and list to lst, as dict and list builtin functions. In addition, there may be a bug in your code, as you are never actually using item; I'm not sure what it's purpose in your code was.

answered Jul 18, 2015 at 21:28

kirbyfan64sos

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Laszlowaty · Accepted Answer · 2015-07-18 21:24:14Z

0

This should be helpfull:

list = ["item1", "item2", "item3"]

for i in xrange(len(list)): # len(list) returns length of the list
    print(list[i])

answered Jul 18, 2015 at 21:24

Laszlowaty

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2 Comments

kirbyfan64sos Over a year ago

xrange is Python 2-only.

Laszlowaty Over a year ago

There is a python-2.7 tag in this question ;-)

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