I have a list which has number and description alternatively. I would like to split it into 2 based on the index so that I can separate number and description. I tried this but it doesn't work-
list to be split based on index is named list2
for i in range (0,len(list2),1):
if (i%2==0):
new_list1=list() #list only with values
dummy=list2[i]
i1=0
new_list1[i1]=dummy
i1=i1+2
else:
new_lisy2=list() #list with description only
dummy2=list2[i]
i2=1
new_list2[i2]=dummy2
i2=i2+2
This should work but sadly it doesn't and when I debug it skips going to else statement. Please help.
I have a list which has number and description alternatively
Why not just use slicing:
original_list = [1,'one',2,'two',3,'three']
numbers_list = a[::2] # start at index[0], continue through list, get every 2nd element
>>> print numbers_list
[1, 2, 3]
strings_list = a[1::2] # start at index[1], continue through list, get every 2nd element
>>> print strings_list
['one', 'two', 'three']
Related
I'm having trouble converting my working code from lists to dictionaries. The basics of the code checks a file name for any keywords within the list.
But I'm having a tough time understanding dictionaries to convert it. I am trying to pull the name of each key and compare it to the file name like I did with lists and tuples. Here is a mock version of what i was doing.
fname = "../crazyfdsfd/fds/ss/rabbit.txt"
hollow = "SFV"
blank = "2008"
empty = "bender"
# things is list
things = ["sheep", "goat", "rabbit"]
# other is tuple
other = ("sheep", "goat", "rabbit")
#stuff is dictionary
stuff = {"sheep": 2, "goat": 5, "rabbit": 6}
try:
print(type(things), "things")
for i in things:
if i in fname:
hollow = str(i)
print(hollow)
if hollow == things[2]:
print("PERFECT")
except:
print("c-c-c-combo breaker")
print("\n \n")
try:
print(type(other), "other")
for i in other:
if i in fname:
blank = str(i)
print(blank)
if blank == other[2]:
print("Yes. You. Can.")
except:
print("THANKS OBAMA")
print("\n \n")
try:
print(type(stuff), "stuff")
for i in stuff: # problem loop
if i in fname:
empty = str(i)
print(empty)
if empty == stuff[2]: # problem line
print("Shut up and take my money!")
except:
print("CURSE YOU ZOIDBERG!")
I am able to get a full run though the first two examples, but I cannot get the dictionary to run without its exception. The loop is not converting empty into stuff[2]'s value. Leaving money regrettably in fry's pocket. Let me know if my example isn't clear enough for what I am asking. The dictionary is just short cutting counting lists and adding files to other variables.
A dictionary is an unordered collection that maps keys to values. If you define stuff to be:
stuff = {"sheep": 2, "goat": 5, "rabbit": 6}
You can refer to its elements with:
stuff['sheep'], stuff['goat'], stuff['rabbit']
stuff[2] will result in a KeyError, because the key 2 is not found in your dictionary. You can't compare a string with the last or 3rd value of a dictionary, because the dictionary is not stored in an ordered sequence (the internal ordering is based on hashing). Use a list or tuple for an ordered sequence - if you need to compare to the last item.
If you want to traverse a dictionary, you can use this as a template:
for k, v in stuff.items():
if k == 'rabbit':
# do something - k will be 'rabbit' and v will be 6
If you want to check to check the keys in a dictionary to see if they match part of a string:
for k in stuff.keys():
if k in fname:
print('found', k)
Some other notes:
The KeyError would be much easier to notice... if you took out your try/except blocks. Hiding python errors from end-users can be useful. Hiding that information from YOU is a bad idea - especially when you're debugging an initial pass at code.
You can compare to the last item in a list or tuple with:
if hollow == things[-1]:
if that is what you're trying to do.
In your last loop: empty == str(i) needs to be empty = str(i).
In short i want to number chars in order from 1 to n without changing the position of the chars in a list.
Suppose I have a list called key = ['c', 'a', 't'] How would i go about
assigning a number to each letter depending on where it is situated in the alphabet with respect to the other letters. Starting at 1 and going until len(key) such that our key becomes [ 2, 1, 3]
I'm really stumped. I have a way to convert them to numbers but very unsure as to how to compare them such that the above happens any help, tips, ideas or explanations would be appreciated.
this is what i have so far...
import string
key = list(input("enter key: ").upper())
num = []
for i in key:
num.append(string.ascii_uppercase.index(i)+1)
This solution assumes that duplicate entries should be assigned the same number, so that
# ['c','a','t'] -> [2, 1, 3]
# ['c','a','t','c','a','t'] -> [2, 1, 3, 2, 1, 3]
You can write a simple function like this:
def get_alphabet_pos(lst):
uniques = sorted(set(lst)) # set() to filter uniques, then order by value
numbers = {letter: i+1 for i, letter in enumerate(uniques)} # build a lookup dict
return [numbers[key] for key in lst]
get_alphabet_pos('cat') # [2, 1, 3]
So here's what happens in the function:
In line 1 of the function we convert your list to a set to remove any duplicate values. From the docs # https://docs.python.org/3/tutorial/datastructures.html#sets:
A set is an unordered collection with no duplicate elements.
Still in line 1 we sort the set and convert it back into a list. Thanks to #StefanPochmann for pointing out that sorted() takes care of the list conversion.
In line 2, we use enumerate() so we can iterate over the indices and values of our list of unique values: https://docs.python.org/3/library/functions.html#enumerate
The rest of line 2 is a simple dict comprehension to build a dictionary of letter -> number mappings. We use the dictionary in line 3 to look up the numbers for each letter in our input dict.
You might have to modify this slightly depending on how you want to handle duplicates :)
I have a dictionary that looks like:
dictionary = {'article1.txt': {'harry': 3, 'hermione': 2, 'ron': 1},
'article2.txt': {'dumbledore': 1, 'hermione': 3},
'article3.txt': {'harry': 5}}
And I'm interested in picking the article with the most number of occurences of Hermione. I already have code that selects the outer keys (article1.txt, article2.txt) and inner key hermione.
Now I want to be able to have code that sorts the dictionary into a list of ascending order for the highest number occurrences of the word hermione. In this case, I want a list such that ['article1.txt', 'article2.txt']. I tried it with the following code:
#these keys are generated from another part of the program
keys1 = ['article1.txt', 'article2.txt']
keys2 = ['hermione', 'hermione']
place = 0
for i in range(len(keys1)-1):
for j in range(len(keys2)-1):
if articles[keys1[i]][keys2[j]] > articles[keys1[i+1]][keys2[j+1]]:
ordered_articles.append(keys1[i])
place += 1
else:
ordered_articles.append(place, keys1[i])
But obviously (I'm realizing now) it doesn't make sense to iterate through the keys to check if dictionary[key] > dictionary[next_key]. This is because we would never be able to compare things not in sequence, like dictionary[key[1]] > dictionary[key[3]].
Help would be much appreciated!
It seems that what you're trying to do is sort the articles by the amount of 'hermiones' in them. And, python has a built-in function that does exactly that (you can check it here). You can use it to sort the dictionary keys by the amount of hermiones each of them points to.
Here's a code you can use as example:
# filters out articles without hermione from the dictionary
# value here is the inner dict (for example: {'harry': 5})
dictionary = {key: value for key, value in dictionary.items() if 'hermione' in value}
# this function just returns the amount of hermiones in an article
# it will be used for sorting
def hermione_count(key):
return dictionary[key]['hermione']
# dictionary.keys() is a list of the keys of the dictionary (the articles)
# key=... here means we use hermione_count as the function to sort the list
article_list = sorted(dictionary.keys(), key=hermione_count)
If I have a list:
list = ('john', 'adam', 'tom', 'danny')
and I want a sorted output with the items where the first letter is between 'a' and 'h', like:
('adam', 'danny', 'john')
which sorting function in Python do I need to complete this task?
This is the code i tried:
l = list()
while True:
s = raw_input("Enter a username: ")
l.append(s)
print sorted(l)
You need 2 distinct things: a list with just the elements that begin with an acceptable letter, and then the sorted version of that list. The former can be done with a list comprehension (although, as #jonrsharpe points out, you look like you want tuples, so you meat need to convert to a list & the convert the result back).
I have a text file like:
abc
abc
abc
def
def
def
...
...
...
...
Now I would like o create a list
list1=['abc','abc','abc']
list2=['def','def','def']
....
....
....
I would like to know how to check if next element is similar to previous element in a python for loop.
You can create a list comprehension and check if the ith element is equal to the ith-1 element in your list.
[ list1[i]==list1[i-1] for i in range(len(list1)) ]
>>> list1=['abc','abc','abc']
>>> [ list1[i]==list1[i-1] for i in range(len(list1)) ]
[True, True, True]
>>> list1=['abc','abc','abd']
>>> [ list1[i]==list1[i-1] for i in range(len(list1)) ]
[False, True, False]
This can be written within a for loop as well:
aux_list = []
for i in range(len(list1)):
aux_list.append(list1[i]==list1[i-1])
Check this post:
http://www.pythonforbeginners.com/lists/list-comprehensions-in-python/
for i in range(1,len(list)):
if(list[i] == list[i-1]):
#Over here list[i] is equal to the previous element i.e list[i-1]
file = open('workfile', 'r') # open the file
splitStr = file.read().split()
# will look like splitStr = ['abc', 'abc', 'abc', 'def', ....]
I think the best way to progress from here would be to use a dictionary
words = {}
for eachStr in splitStr:
if (words.has_key(eachStr)): # we have already found this word
words[eachStr] = words.get(eachStr) + 1 # increment the count (key) value
else: # we have not found this word yet
words[eachStr] = 1 # initialize the new key-value set
This will create a dictionary so the result would look like
print words.items()
[('abc', 3), ('def', 3)]
This way you store all of the information you want. I proposed this solution because its rather messy to create an unknown number of lists to accommodate what you want to do, but it is easy and memory efficient to store the data in a dictionary from which you can create a list if need be. Furthermore, using dictionaries and sets allow you to have a single copy of each string (in this case).
If you absolutely need new lists let me know and I will try to help you figure it out