Right, I have a dictionary like this one:
my_dict = {'BAM': (1.985, 1.919), 'PLN': (4.509, 4.361),'SEK': (9.929, 9.609), 'CZK': (27.544, 26.544),
'NOK': (9.2471, 8.9071), 'AUD': (1.4444, 1.4004),
'HUF': (315.89, 307.09), 'GBP': (0.8639, 0.8399),
'HRK': (7.6508, 7.4208), 'RUB': (71.9393, 66.5393),
'USD': (1.0748, 1.0508), 'MKD': (62.11, 60.29),
'CHF': (1.0942, 1.0602), 'JPY': (121.83, 118.03),
'BGN': (1.979, 1.925), 'RSD': (124.94, 121.14),
'DKK': (7.5521, 7.3281), 'CAD': (1.4528, 1.4048)}
I need to write a function (my_dict, ["GBP", "USD", "RUB", "HRK", "HUF"]) that returns this:
GBP......0.8639......0.8399
USD......1.0748......1.0508
RUB.....71.9393.....66.5393
HRK......7.6508......7.4208
HUF....315.8900....307.0900
We are learning how to format string, and I have no idea how to approach this. Any help would be appreciated. It needs to be formatted exactly like this, with all the dots and stuff (without the empty space before GBP).
One of the simpler ways to do what you're asking is to use a
list comprehension:
def my_function(dict, list):
return ["{0}......{1}......{2}".format(item, dict[item][0], dict[item][1])
for item in list
if item in dict]
my_function will return a list of currency......value......value items. In fact, you don't even need to create a function:
strings = ["{0}......{1}......{2}".format(item, dict[item][0], dict[item][1])
for item in list
if item in dict]
If however you don't want to use a list comprehension, the same function could look like this:
def my_function(dict, list):
strings = []
for item in list:
if item in dict:
strings.append("{0}......{1}......{2}".format(item, dict[item][0], dict[item][1]))
return strings
Related
I have a list of strings. Each string has the same length/number of characters in the format
xyzw01.ext or xyzv02.ext, etc.
For example
list 1: ['ABCJ01.ext','CDEJ02.ext','ADEJ01.ext','CDEJ01.ext','ABCJ02.ext','CDEJ03.ext']
list 2: ['ABCJ01.ext','ADEJ01.ext','CDEJ01.ext','RPNJ01.ext','PLEJ01.ext']
I would like from these lists to build new lists with only the strings with highest number.
So from list 1 I would like to get
['ADEJ01.ext','ABCJ02.ext','CDEJ03.ext']
while for list 2 I would like to get the same list since all numbers are 01.
Is there a "simple" way of achieving this?
You can use defaultdict and max
from collections import defaultdict
def fun(lst):
res = defaultdict(list)
for x in lst:
res[x[:4]].append(x)
return [max(res[x], key=lambda x: x[4:6]) for x in res]
lst = ['ABCJ01.ext','CDEJ02.ext','ADEJ01.ext','CDEJ01.ext','ABCJ02.ext','CDEJ03.ext']
lst2 = ['ABCJ01.ext','ADEJ01.ext','CDEJ01.ext','RPNJ01.ext','PLEJ01.ext']
print(fun(lst))
print(fun(lst2))
Output:
['ABCJ02.ext', 'CDEJ03.ext', 'ADEJ01.ext']
['ABCJ01.ext', 'ADEJ01.ext', 'CDEJ01.ext', 'RPNJ01.ext', 'PLEJ01.ext']
The easiest way is probably to use an intermediate data structure, like a dict - sort the list items into buckets based on the first part of their names, and then take the maximum number for each bucket. We can just use the built-in max() without a key, since as-given lexicographic sorting works to find the largest. If that's not sufficient, you could use more regex to take the number out of the item and use it as the key instead.
import re
def filter_list(lst):
prefixes = {}
for item in lst:
# use regex to isolate the non-numeric characters at the start of the string
prefix = re.match(r'^([^0-9]*)', item).group(1)
# make a bucket based on each prefix, and put the item in it
prefixes.setdefault(prefix, [])
prefixes[prefix].append(item)
# make a list comprehension taking the maximum item from each bucket
return [max(value) for value in prefixes.values()]
>>> a = ['ABCJ01.ext','CDEJ02.ext','ADEJ01.ext','CDEJ01.ext','ABCJ02.ext','CDEJ03.ext']
>>> b = ['ABCJ01.ext','ADEJ01.ext','CDEJ01.ext','RPNJ01.ext','PLEJ01.ext']
>>> filter_list(a)
['ABCJ02.ext', 'CDEJ03.ext', 'ADEJ01.ext']
>>> filter_list(b)
['ABCJ01.ext', 'ADEJ01.ext', 'CDEJ01.ext', 'RPNJ01.ext', 'PLEJ01.ext']
In python 3.7+, this should preserve the order of list from the first occurrence of each prefix (i.e. CDEJ03.ext will precede ADEJ01.ext in the output because CDEJ02.ext precedes it in the input).
To get the output in the exact same order as the original list, behavior, you'd want to explicitly reassign the key instead of using .setdefault(), perhaps with a pattern like prefixes[prefix] = prefixes[prefix] if prefix in prefixes else [].
Maybe it is ordinary issue regarding iterating thru a dict. Please find below imovel.txt file, whose content is as follows:
{'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
As you can see this is not a ordinary dictionary, with a key value pair; but a key with a list as key and another list as value
My code is:
#/usr/bin/python
def load_dict_from_file():
f = open('../txt/imovel.txt','r')
data=f.read()
f.close()
return eval(data)
thisdict = load_dict_from_file()
for key,value in thisdict.items():
print(value)
and yields :
['primeiro', 'segundo', 'terceiro'] ['101', '201', '301']
I would like to print a key,value pair like
{'primeiro':'101, 'segundo':'201', 'terceiro':'301'}
Given such txt file above, is it possible?
You should use the builtin json module to parse but either way, you'll still have the same structure.
There are a few things you can do.
If you know both of the base key names('Andar' and 'Apto') you can do it as a one line dict comprehension by zipping the values together.
# what you'll get from the file
thisdict = {'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
# One line dict comprehension
newdict = {key: value for key, value in zip(thisdict['Andar'], thisdict['Apto'])}
print(newdict)
If you don't know the names of the keys, you could call next on an iterator assuming they're the first 2 lists in your structure.
# what you'll get from the file
thisdict = {'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
# create an iterator of the values since the keys are meaningless here
iterator = iter(thisdict.values())
# the first group of values are the keys
keys = next(iterator, None)
# and the second are the values
values = next(iterator, None)
# zip them together and have dict do the work for you
newdict = dict(zip(keys, values))
print(newdict)
As other folks have noted, that looks like JSON, and it'd probably be easier to parse it read through it as such. But if that's not an option for some reason, you can look through your dictionary this way if all of your lists at each key are the same length:
for i, res in enumerate(dict[list(dict)[0]]):
ith_values = [elem[i] for elem in dict.values()]
print(ith_values)
If they're all different lengths, then you'll need to put some logic to check for that and print a blank or do some error handling for looking past the end of the list.
list = [1,2,"three",{"number":4}]
for item in list:
if type(item) is dict:
print([val for val in item.values()][0])
else:
print(item)
In the example above, provided I would not know that the key of the dictionary item is named number, is there an easier (or more elegant) way to print just the value of the item?
The arguably more elegant way to extract the first value from a dict would be to write:
print(next(iter(item.values())))
My dict (cpc_docs) has a structure like
{
sym1:[app1, app2, app3],
sym2:[app1, app6, app56, app89],
sym3:[app3, app887]
}
My dict has 15K keys and they are unique strings. Values for each key are a list of app numbers and they can appear as values for more than one key.
I've looked here [Python: Best Way to Exchange Keys with Values in a Dictionary?, but since my value is a list, i get an error unhashable type: list
I've tried the following methods:
res = dict((v,k) for k,v in cpc_docs.items())
for x,y in cpc_docs.items():
res.setdefault(y,[]).append(x)
new_dict = dict (zip(cpc_docs.values(),cpc_docs.keys()))
None of these work of course since my values are lists.
I want each unique element from the value lists and all of its keys as a list.
Something like this:
{
app1:[sym1, sym2]
app2:[sym1]
app3:[sym1, sym3]
app6:[sym2]
app56:[sym2]
app89:[sym2]
app887:[sym3]
}
A bonus would be to order the new dict based on the len of each value list. So like:
{
app1:[sym1, sym2]
app3:[sym1, sym3]
app2:[sym1]
app6:[sym2]
app56:[sym2]
app89:[sym2]
app887:[sym3]
}
Your setdefault code is almost there, you just need an extra loop over the lists of values:
res = {}
for k, lst in cpc_docs.items():
for v in lst:
res.setdefault(v, []).append(k)
First create a list of key, value tuples
new_list=[]
for k,v in cpc_docs.items():
for i in range(len(v)):
new_list.append((k,v[i]))
Then for each tuple in the list, add the key if it isn't in the dict and append the
doc_cpc = defaultdict(set)
for tup in cpc_doc_list:
doc_cpc[tup[1]].add(tup[0])
Probably many better ways, but this works.
I am an extremely begginer learning python to tackle some biology problems, and I came across lists and its various methods. Basically, when I am running print to my variable I get None as return.
Example, trying to print a sorted list assigned to a variable
list1=[1,3,4,2]
sorted=list1.sort()
print(sorted)
I receive None as return. Shouldn't this provide me with [1,2,3,4]
However, when printing the original list variable (list1), it gives me the sorted list fine.
Because the sort() method will always return None. What you should do is:
list1=[1,3,4,2]
list1.sort()
print(list1)
Or
list1=[1,3,4,2]
list2 = sorted(list1)
print(list2)
You can sort lists in two ways. Using list.sort() and this will sort list, or new_list = sorted(list) and this will return a sorted list new_list and list will not be modified.
So, you can do this:
list1=[1,3,4,2]
sorted=sorted(list1)
print(sorted)
Or you can so this:
list1=[1,3,4,2]
list1.sort()
print(list1)