Python Iterating through List of List

Question

Heres my code

stockList = [
    ['AMD', '57.00', '56.23', '58.40', '56.51'],
    ['AMZN', '3,138.29', '3,111.03', '3242.56689', '3,126.58'],
    ['ATVI', '80.76', '79.16', '81.86', '79.55'],
    ['BA', '178.63', '168.86', '176.96', '169.70'],
    ['BAC', '24.42', '23.43', '23.95', '23.54'],
    ['DAL', '26.43', '25.53', '26.87', '25.66'],
    ['FB', '241.75', '240.00', '248.06', '241.20'],
    ['GE', '7.04', '6.76', '6.95', '6.79'],
    ['GOOGL', '1,555.92', '1,536.36', '1,576.03', '1,544.04'],
    ['GPS', '12.77', '12.04', '12.72', '12.10'],
    ['GRUB', '70.96', '69.71', '70.65', '70.06'],
    ['HD', '262.42', '258.72', '261.81', '260.01'],
    ['LUV', '33.62', '32.45', '33.53', '32.61'],
    ['MSFT', '208.75', '206.72', '213.58', '207.76'],
    ['MU', '51.52', '50.49', '52.31', '50.74'],
    ['NFLX', '490.10', '492.26', '511.52', '494.72', 'SUCCESS'],
    ['PCG', '9.49', '8.96', '9.52', '9.01'],
    ['PFE', '36.69', '35.87', '37.02', '36.05'],
    ['QQQ', '264.00', '263.27', '267.11', '264.58', 'SUCCESS'],
    ['ROKU', '153.36', '148.37', '153.70', '149.11'],
    ['SHOP', '952.83', '976.45', '1,036.25', '981.33', 'SUCCESS'],
    ['SPY', '325.01', '323.64', '325.47', '325.25', 'SUCCESS'],
    ['SQ', '126.99', '125.13', '130.80', '125.76'],
    ['T', '30.25', '29.58', '30.07', '29.73'],
    ['TSLA', '1,568.36', '1,646.56', '1,712.58', '1,654.79', 'SUCCESS'],
    ['TTWO', '153.06', '152.45', '154.47', '153.22', 'SUCCESS'],
    ['TWTR', '37.01', '36.03246', '36.7210083', '36.21'],
    ['WFC', '26.20', '24.45272', '25.0438213', '24.57'],
    ['WMT', '132.33', '130.8515', '132.522049', '131.51']
]

keyword = 'SUCCESS'
secondList = []
for item in stockList:
    if item[4] == keyword:
        secondList.append(stockList[0])
print(secondList)

My use case is, to go through this lists of list, find which list contains the keyword, from there send the first item in the list. I am able to get it with one single list, however I can't do it with a list of list.

On top of that, how would I go through a dictionary containing lists?

{
    'majorDimension': 'ROWS',
    'range': 'Sheet1!A2:F30',
    'values': [
        ['AMD', '57.00', '56.23', '58.40', '56.51'],
        ['AMZN', '3,138.29', '3,111.03', '3242.56689', '3,126.58'],
        ['ATVI', '80.76', '79.16', '81.86', '79.55'],
        ['BA', '178.63', '168.86', '176.96', '169.70'],
        ['BAC', '24.42', '23.43', '23.95', '23.54'],
        ['DAL', '26.43', '25.53', '26.87', '25.66'],
        ['FB', '241.75', '240.00', '248.06', '241.20'],
        ['GE', '7.04', '6.76', '6.95', '6.79'],
        ['GOOGL', '1,555.92', '1,536.36', '1,576.03', '1,544.04'],
        ['GPS', '12.77', '12.04', '12.72', '12.10'],
        ['GRUB', '70.96', '69.71', '70.65', '70.06'],
        ['HD', '262.42', '258.72', '261.81', '260.01'],
        ['LUV', '33.62', '32.45', '33.53', '32.61'],
        ['MSFT', '208.75', '206.72', '213.58', '207.76'],
        ['MU', '51.52', '50.49', '52.31', '50.74'],
        ['NFLX', '490.10', '492.26', '511.52', '494.72', 'SUCCESS'],
        ['PCG', '9.49', '8.96', '9.52', '9.01'],
        ['PFE', '36.69', '35.87', '37.02', '36.05'],
        ['QQQ', '264.00', '263.27', '267.11', '264.58', 'SUCCESS'],
        ['ROKU', '153.36', '148.37', '153.70', '149.11'],
        ['SHOP', '952.83', '976.45', '1,036.25', '981.33', 'SUCCESS'],
        ['SPY', '325.01', '323.64', '325.47', '325.25', 'SUCCESS'],
        ['SQ', '126.99', '125.13', '130.80', '125.76'],
        ['T', '30.25', '29.58', '30.07', '29.73'],
        ['TSLA', '1,568.36', '1,646.56', '1,712.58', '1,654.79', 'SUCCESS'],
        ['TTWO', '153.06', '152.45', '154.47', '153.22', 'SUCCESS'],
        ['TWTR', '37.01', '36.03246', '36.7210083', '36.21'],
        ['WFC', '26.20', '24.45272', '25.0438213', '24.57'],
        ['WMT', '132.33', '130.8515', '132.522049', '131.51'],
    ]
}

Edit: I was able to get it with keyword = 'SUCCESS' for item in stockList: if item[-1] == keyword: print(item) — JaydaMan
– JaydaMan, Commented Jul 22, 2020 at 5:00
Hey Jayda, please check out my answer, and let me know, if that helped you. — Alok
– Alok, Commented Jul 22, 2020 at 5:49
So I actually had some time to look at it again, this is what I came up with: Can I get some feedback? I realized I could turn the dictionary into a list since I was unaware of how to access the list part of the dictionary. If that makes sense, still new to python stockList = list(response.values()) keyword = 'SUCCESS' for item in stockList: for lul in item: if lul[-1] == keyword: print(lul[0]). I figured I had to have 2 loops since it is a nested nested list? — JaydaMan
– JaydaMan, Commented Jul 22, 2020 at 6:05
See, I am a bit confused with the requirement, but if you really need a suggestion on your new idea based upon the new values, showing the new value, will enable people to have more clarity or you can ask a new question with your try in that. People would definitely suggest you some best approach to deal with the queston. Make sure you add your work on the new value, if you are asking a new question. — Alok
– Alok, Commented Jul 22, 2020 at 6:23

Sam Morgan · Accepted Answer · 2020-07-22 05:23:26Z

1

List comprehension makes this pretty simple. Try the following:

keyword = "SUCCESS"

# PEP8 calls for lower_underscore_case here
second_list = [i[0] for i in stockList if keyword in i]

print(second_list)

For the proposed dictionary structure, you'd just access the key containing the list, since not every value in that dict is a list:

second_list = [i[0] for i in stockList["values"] if keyword in i]

answered Jul 22, 2020 at 5:23

Sam Morgan

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Comments

halfer · Accepted Answer · 2020-07-30 11:07:47Z

0

Based upon your question understanding. Your question is divided into two parts, these are:

How to iterate over list of lists, and get the first item from the nested list, and store it in another list
How to iterate over dictionary item, to perform the same operation

If my understanding is right, then you might want to check this out.

Please note: I have not used variable keyword, simply used "SUCCESS", just replace keyword with "SUCCESS" in the code, and you are good to go.

1. FIRST SOLUTION

# to get nested list
for item in stockList:
    # this checks whether SUCCESS is present inside a list
    # python way of doing it
    if "SUCCESS" in item: secondList.append(item[0])
    
print(secondList)

# OUTPUT 
# >>> ['NFLX', 'QQQ', 'SHOP', 'SPY', 'TSLA', 'TTWO']

OR

You can do this in more pythonic way, that is to use List Comprehension

# single line approach, getting the same result
secondList = [item[0] for item in stockList if "SUCCESS" in item]
    
print(secondList)

# OUTPUT
# >>> ['NFLX', 'QQQ', 'SHOP', 'SPY', 'TSLA', 'TTWO']

2. SECOND SOLUTION

In order to get the result, first you need to assign the Dictionary to your variable, in my case, I have assigned to a variable called stockListDictionary

secondList = []        
# to get a value from key specifically
# likt any dictionary key dictionary["key_name"]
for item in stockListDictionary["values"]:
    if "SUCCESS" in item: secondList.append(item[0])

print(secondList)

# OUTPUT
# >>> ['NFLX', 'QQQ', 'SHOP', 'SPY', 'TSLA', 'TTWO']

OR

Using List Comprehension

secondList = [item[0] for item in stockListDictionary["values"] if "SUCCESS" in item]

print(secondList)

# OUTPUT
# >>> ['NFLX', 'QQQ', 'SHOP', 'SPY', 'TSLA', 'TTWO']

edited Jul 30, 2020 at 11:07

halfer

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answered Jul 22, 2020 at 5:48

Alok

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2 Comments

JaydaMan Over a year ago

thanks for the great explanations man, I ended up solving it with another solution two loops

Alok Over a year ago

Great @JaydaMan, our main motive is to help every one for a kick start, and then every one can blast off at their own pace. BTW - always develop or code something keeping the time complexity in mind. The better the solution is, the better you grow. For example using two loops is O(N^2), which my solution is also. But always try to make the solution fall under the best complexity, that is: O(1), O(NlogN), O(N).

O.rka · Accepted Answer · 2020-07-22 05:51:46Z

0

What about something like this?

keywords={"SUCCESS"}
d = # the dictionary
second_list = list()
for nested_lists in d["values"]:
    for stock_info in nested_lists:
        stock_ticker = stock_info[0]
        if stock_ticker in keywords:
            info = set(stock_info[1:])
            if info & keywords:
                second_list.append(stock_ticker)

Is this better? It should allow you to have more than one keyword.

edited Jul 22, 2020 at 5:51

answered Jul 22, 2020 at 4:42

O.rka

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1 Comment

JaydaMan Over a year ago

I guess first part is, given the first list, how can I iterate through every list and see which lists have Success. For example given the original stocklist, I want the code to return NFLX,QQQ,SHOP,SPY,TSLA,TTWO since they have 'SUCCESS' in their list. if that makes sense?

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