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Convert web scraped string list to formatted CSV


By : jasonseiler
Date : October 17 2020, 08:10 PM
Hope this helps Your assumptions are correct! There's likely a more efficient way to do this, but this requires very few code changes.
Split the string using
code :
split_name = name.split("\n")
no_blanks = [ x for x in split_name if len(x) > 0 ]
with open('index.csv', 'a') as csv_file:
writer = csv.writer(csv_file)
line = []
for i in range(len(no_blanks)):
    line.append(no_blanks[I].strip())
    if len(line) == 8:
        writer.writerow(line)
        line = []


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Python convert formatted string to list


By : Ulrich Schonhardt
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I have a string "[u'foo']" (Yes, it includes the square brackets and the u''). I have to convert that to a list which looks like [u'foo'].
code :
>>> import ast
>>> s = "[u'foo']"
>>> ast.literal_eval(s)
[u'foo']

Back-convert a list (formatted as a string) into a list


By : M.Davydov
Date : March 29 2020, 07:55 AM
Any of those help I have a list of dictionaries, it looks something like this: , Use ast.literal_eval to convert a string to a python object(safely):
code :
>>> from ast import literal_eval
>>> strs = "[{'T': 13472}, {'A': 13472}, {'C': 9866, 'T': 3606}, {'G': 13472}, {'G': 13472}, {'A': 221, 'C': 26, 'T': 12845, 'G': 380}, {'T': 13472}, {'A': 13472}, {'C': 546, 'T': 12926}, {'C': 13472}, {'A': 13472}, {'C': 10674, 'T': 2798}, {'C': 13472}, {'A': 13472}, {'C': 554, 'T': 12918}, {'C': 13472}, {'A': 13472}]"
>>> literal_eval(strs)
[{'T': 13472}, {'A': 13472}, {'C': 9866, 'T': 3606}, {'G': 13472}, {'G': 13472}, {'A': 221, 'C': 26, 'T': 12845, 'G': 380}, {'T': 13472}, {'A': 13472}, {'C': 546, 'T': 12926}, {'C': 13472}, {'A': 13472}, {'C': 10674, 'T': 2798}, {'C': 13472}, {'A': 13472}, {'C': 554, 'T': 12918}, {'C': 13472}, {'A': 13472}]

Convert list in Scala to a formatted string


By : Cliff Wang
Date : March 29 2020, 07:55 AM
This might help you Have a look at mkString. In short:
code :
scala> List(1,2,3).mkString("/")
res0: String = 1/2/3

scala> List(1,2,3).mkString
res0: String = 123

// def mkString(start: String,sep: String,end: String): String 
scala> List(1,2,3).mkString("@", "/", "@")
res1: String = @1/2/3@

Unable to Convert scraped data from list to regular string


By : user6845114
Date : March 29 2020, 07:55 AM
wish help you to fix your issue When I run my crawler it fetches the results as list. However, I expected to have that in regular string being displayed in two columns. for any suggestion. , Try iterating over both list sequentially, like this:
code :
import requests
from lxml import html

url="http://www.wiseowl.co.uk/videos/"
def Startpoint(links):
    response = requests.get(links)
    tree = html.fromstring(response.text)
    Title= tree.xpath("//p[@class='woVideoListDefaultSeriesTitle']/a/text()")
    Link=tree.xpath("//p[@class='woVideoListDefaultSeriesTitle']/a/@href")
    for i,j in zip(Title, Link):
        print('{:<70}{}'.format(i,j))

Startpoint(url)

Convert List enumeration to formatted string


By : bontey
Date : March 29 2020, 07:55 AM
like below fixes the issue just use declarative IO, i.e. express your code in a DCG (Definite Clause Grammar):
code :
:- use_module(library(clpfd)).

tree(0,[]) --> "".
tree(N, [Lt, Rt]) -->
    {N #> 0,
     N #= N1 + N2 + 1,
     N1 #>= 0, N2 #>= 0
    },
    "(", tree(N1, Lt), ".", tree(N2, Rt), ")".
?- phrase(tree(2,T),S),format('~s~n',[S]).
(.(.))
T = [[], [[], []]],
S = [40, 46, 40, 46, 41, 41] ;
((.).)
T = [[[], []], []],
S = [40, 40, 46, 41, 46, 41] ;
false.
:- set_prolog_flag(double_quotes, chars).
?- phrase(tree(2,T),S).
T = [[], [[], []]],
S = ['(', '.', '(', '.', ')', ')'] ;
T = [[[], []], []],
S = ['(', '(', '.', ')', '.', ')'] ;
false.
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