Stack Overflow Asked by JaydaMan on November 18, 2021
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'],
]
}
Based upon your question understanding. Your question is divided into two parts, these are:
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']
Answered by Alok on November 18, 2021
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 by Sam Morgan on November 18, 2021
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.
Answered by O.rka on November 18, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP