Sorry for the poor title, basically I want to access a list item based on another list item:
catList = [john, james, jack]
dateList = [1y, 2y, 3m]
for item in catList:
if item in typeList[1]:
calendar += 1
if item in typeList[2]:
connector += 1
print dateList[item]
It's the last line that I want to achieve, essentially to print the value of dateList which corresponds to the value of catList.
Any ideas? Thanks
Edit: My catDict & dateDict are lists so they don't have keys, just values.
If the keys to catDict are numeric strings you can use print dateDict[int(item)]. Otherwise, I'm uncertain - we need to see what the declaration of catDict looks like.
If you're using lists you want to do this:
for idx, val in enumerate(catList):
print dateList[idx]
Related
New to python and for this example list
lst = ['<name>bob</name>', '<job>doctor</job>', '<gender>male</gender>', '<name>susan</name>', '<job>teacher</job>', '<gender>female</gender>', '<name>john</name>', '<gender>male</gender>']
There are 3 categories of name, job, and gender. I would want those 3 categories to be on the same line which would look like
<name>bob</name>, <job>doctor</job>, <gender>male</gender>
My actual list is really big with 10 categories I would want to be on the same line. I am also trying to figure out a way where if one of the categories is not in the list, it would print something like N/A to indicate that it is not in the list
for example I would want it to look like
<name>bob</name>, <job>doctor</job>, <gender>male</gender>
<name>susan</name>, <job>teacher</job>, <gender>female</gender>
<name>john</name>, N/A, <gender>male</gender>
What would be the best way to do this?
This is one way to do it. This would handle any length list, and guarantee grouping no matter how long the lists are as long as they are in the correct order.
Updated to convert to dict, so you can test for key existence.
lst = ['<name>bob</name>', '<job>doctor</job>', '<gender>male</gender>', '<name>susan</name>', '<job>teacher</job>', '<gender>female</gender>', '<name>john</name>', '<gender>male</gender>']
newlst = []
tmplist = {}
for item in lst:
value = item.split('>')[1].split('<')[0]
key = item.split('<')[1].split('>')[0]
if '<name>' in item:
if tmplist:
newlst.append(tmplist)
tmplist = {}
tmplist[key] = value
#handle the remaining items left over in the list
if tmplist:
newlst.append(tmplist)
print(newlst)
#test for existance
for each in newlst:
print(each.get('job', 'N/A'))
I'm new to python and I was wondering if there's a way for me to pull a value at a specific index. Let's say I have a key with multiple values(list) associated with it.
d = {'ANIMAL' : ['CAT','DOG','FISH','HEDGEHOG']}
Let's say I want to iterate through values and print out the value if it's equal to 'DOG'. Do Values, Key pairs have a specific index associated with the position of Values?
I've try reading up on dict and how it works apparently you can't really index it. I just wanted to know if there's a way to get around that.
You can perform the following (comments included):
d = {'ANIMAL' : ['CAT','DOG','FISH','HEDGEHOG']}
for keys, values in d.items(): #Will allow you to reference the key and value pair
for item in values: #Will iterate through the list containing the animals
if item == "DOG":
print(item)
print(values.index(item)) #will tell you the index of "DOG" in the list.
So maybe this will help:
d = {'ANIMAL' : ['CAT','DOG','FISH','HEDGEHOG']}
for item in d:
for animal in (d[item]):
if animal == "DOG":
print(animal)
Update -What if I want to compare the string to see if they're equal or not... let say if the value at the first index is equal to the value at the second index.
You can use this:
d = {'ANIMAL' : ['CAT','DOG','FISH','HEDGEHOG']}
for item in d:
for animal in (d[item]):
if animal == "DOG":
if list(d.keys())[0] == list(d.keys())[1]:
print("Equal")
else: print("Unequal")
Keys and values in a dictionary are indexed by key and do not have a fixed index like in lists.
However, you can leverage the use of 'OrderedDict' to give an indexing scheme to your dictionaries. It is seldom used, but handy.
That being said, dictionaries in python3.6 are insertion ordered :
More on that here :
Are dictionaries ordered in Python 3.6+?
d = {'animal': ['cat', 'dog', 'kangaroo', 'monkey'], 'flower': ['hibiscus', 'sunflower', 'rose']}
for key, value in d.items():
for element in value:
if element is 'dog':
print(value)
does this help? or, you want to print index of key in dictionary?
I have a Queue (size 30) with lines (got from requests r).
Those lines are each dictionaries, how can I look if a dict contains a key 'example'?
def process_queue(queue):
count = 0
for line in queue.get():
count = count + 1
if 'venue' in dict():
print ('Yes')
else:
print ('No')
if count == 30:
break
If you want to check if a dictionary contains a certain key, you can use
if example in list(yourDict.keys()):
# do what you want here
list(yourDict.keys()) will be a list of all the keys in yourDict, in no particular order (where yourDict is a dictionary. This basically reads like an English sentence: "If example is an item in the list, then do something."
Am I able to slice a list of strings? If it is possible could anyone please tell me how to do it so that I am able to print out a particular string instead of the five that make up the list.
Cheers.
eg.
mylist = ['apples' 'oranges' 'lemons' 'cucumbers' 'bananas']
print 'orange'
** The programming language i am using is python. Every time I code it mylist[2] it comes out as an error. The list I am using is extracting the strings from a html rss feed. Each string is a new news heading. However, even when it updates constantly there are always 5 strings in the list and it tells me list index out of range. But if I just print the entire list it works fine**
#URLS for RSS Feeds
url_national = 'http://feeds.news.com.au/public/rss/2.0/news_national_3354.xml'
url_sport = 'http://feeds.news.com.au/public/rss/2.0/news_sport_3168.xml'
url_world = 'http://feeds.news.com.au/public/rss/2.0/news_theworld_3356.xml'
url_technology = 'http://feeds.news.com.au/public/rss/2.0/news_tech_506.xml'
def headlines (url):
web_page = urlopen(url)
html_code = web_page.read()
web_page.close()
return findall(r'<item><title>([^<]*)</title>', html_code)
#headlines list
list_national = [headlines(url_national)]
list_sport = [headlines(url_sport)]
list_world = [headlines(url_world)]
list_technology = [headlines(url_technology)]
def change_category():
if label_colour.get() == 'n':
changeable_label['text'] = list_national #here I would slice it but it doesn't work
elif label_colour.get() == 's':
changeable_label['text'] = list_sport
elif label_colour.get() =='w':
changeable_label['text'] = list_world
else:
changeable_label['text'] = list_technology
the reason I need to slice it into individual heading is so when the radio button is pressed for my GUI it prints them in a numbered list on the label not all just running on one line next to them - sorry i hope that makes sense
What language are you using here? Usually you can use an index to access a particular entry in a list. For example:
print myList[1]
Commas are missing in your list creation. You have to do it like this:
mylist = ['apples', 'oranges', 'lemons', 'cucumbers', 'bananas']
And you will be able to work with your list
mylist[0] # 'apples'
mylist[-1] # 'bananas'
mylist[2] # 'lemons'
I think the error you are getting is something like this:
mylist = ['apples' 'oranges' 'lemons' 'cucumbers' 'bananas']
print mylist[5]
IndexError: list index out of range
The reason is the elements in a list are indexed from 0 not 1.
The mylist has 5 elements starting from 0 to 4. So when you call print mylist[5] it will definitely give an error as there is no 6th element in the list.
Here is the official doc regarding list please have a look.
I hope it was helpful!
I am trying to create an automated python procedure that uses two separate lists to create a dictionary and so far I am failing. I have two sorted lists where the nth item in the fist list corresponds to the nth item in the second list and I want to combine them into a dictionary.
For example, a subset of the 2 lists are as follows;
name = ['Adam', 'Alfred', 'Amy', 'Andy', 'Bob']
year = [1972, 1968, 1985, 1991, 1989]
I would want my output to be:
birth_years = {'Adam':1972, 'Alfred':1968, 'Amy':1985, 'Andy':1991, 'Bob':1989}
I was trying to do this with a for loop, but I could not get it to work. I appreciate any help.
Use the zip and dict functions to construct a dictionary out of a list of tuples:
birth_years = dict(zip(name, year))
And if you're curious, this would be how I would try to do it with a for loop:
birth_years = {}
for index, n in enumerate(name):
birth_years[n] = years[index]
I think I like the first example more.
birth_years = {}
for i in range(len(name)):
birth_years[name[i]] = year[i]
Try a dictionary comprehension:
birth_years = {nm:year[idx] for idx, nm in enumerate(name)}
Your lists would be better named names and years. You got off to a good start by trying to do it with a for loop. Most practical data processing problems involve error checking, which is just a tad difficult with one-liners. Example:
birth_years = {}
for i, name in enumerate(names):
if name in birth_years:
log_duplicate(i, name, years[i]))
else:
birth_years[name] = years[i]