在Python中使用for结构时获取迭代器值

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)

Answer 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.

Answer 2

This should be helpfull:

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

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