> function getDateShortName(s, shortYear=true) {
... let parts = s.split(/\s+/ig)
...
... if(shortYear) {
..... return parts[0].slice(0, 3) + '\'' + parts[1].slice(2)
..... } else {
..... return parts[0].slice(0, 3) + '\'' + parts[1]
https://stackoverflow.com/a/59004613/6615163
In [1]: import re
In [2]: listA = ['123', '345', '678']
In [3]: listB = ['ABC123', 'CDE455', 'GHK678', 'CGH345']
def get_pattern(n=3, separator=' '):
a = 0
pattern = ''
for i in range(0, n):
for j in range(0, n):
Just to understand where exactly you have to do the modification and what little mistake you did has already been pointed out in the above comments.
> var arr = [ "{\"id\": \"amzn\", \"price\": 1785.6600341796875, \"_attachments\": \"attachments/\"}", "{\"id\": \"msft\", \"price\": 138.42999267578125, \"_attachments\": \"attachments/\"}", "{\"id\": \"googl\", \"price\": 1244.280029296875, \"_attachments\": \"attachments/\"}", "{\"id\": \"ba\", \"price\": 331.05999755859375, \"_attachments\": \"attachments/\"}", "{\"id\": \"air.pa\", \"price\": 122, \"_attachments\": \"attachments/\"}" ]
undefined
>
> item0 = arr[0]
'{"id": "amzn", "price": 1785.6600341796875, "_attachments": "attachments/"}'
>
> item0Obj = JSON.parse(item0)
{ id: 'amzn', price: 1785.6600341796875, _attachments: 'attachments/' }
Check my answer at https://stackoverflow.com/a/58067032/6615163
>>> class TEST:
... def __init__(self, project = 'ABC', scenarios = {'A': [1,2,3], 'B': [4,5,6]}):
... self.project = project
... for feature in scenarios:
>>> import requests
>>>
>>> res = requests.get("https://www.amazon.in/b/ref=s9_acsd_hfnv_hd_bw_b3iMERj_ct_x_ct00_w?_encoding=UTF8&node=10485223031&pf_rd_m=A1VBAL9TL5WCBF&pf_rd_s=merchandised-search-3&pf_rd_r=59M3EPGWMDMASM4XWYPJ&pf_rd_r=59M3EPGWMDMASM4XWYPJ&pf_rd_t=101&pf_rd_p=e2e13b35-daa5-587b-aab3-6e9e9135fd3c&pf_rd_p=e2e13b35-daa5-587b-aab3-6e9e9135fd3c&pf_rd_i=3403856031")
>>>
>>> res.status_code
200
>>>
>>> res.text
In Python, list, dictionary are reference types. So if you will append the same dictionary to a list and if you will make any change to any of the dictionary, it will reflect in all other as well because all are pointing to the same memory location.
>>> l = [{"name": "Raghavendra Thakur", "age": 27}] * 4
>>>
>>> l
[{'name': 'Raghavendra Thakur', 'age': 27}, {'name': 'Raghavendra Thakur', 'age': 27}, {'name': 'Raghavendra Thakur', 'age': 27}, {'name': 'Raghavendra Thakur', 'age': 27}]
>>>
>>> l[0]
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""" | |
{ | |
"created_on" : "14 April 2017", | |
"aim_of_script" : "Python script to find the first repeated character in string", | |
"coded_by" : "Rishikesh Agrawani", | |
"python_version" : "2.7.12", | |
"special" : "As an alternative to the pyscript available at http://www.geeksforgeeks.org/find-the-first-repeated-character-in-a-string/?utm_source=feedburner&utm_medium=email&utm_campaign=Feed%3A+Geeksforgeeks+%28GeeksforGeeks%29" | |
} | |
""" |