I have the following html extract
0.94
I am trying to read the href value ie 0.94.I tried the following :
answer = browser.find_element_by_class_name("res")
print answer
output = answer.get_attribute('data-href')
print output
The Result is as follows:
None
I tried various other methods, using find_element_by_xpath etc,but not able to get the desired value ie. 0.94 (as in this example).
How can I get this value in the shortest way? Thanks in advance
Use getText method if you want to to print 0.94
Related
I'm currently using the following formula:
=IFS(regexmatch(A2,"Malaysia"),
B2-dataset!B3,REGEXMATCH(A2,"Saudi Arabia"),
B2-dataset!B7,REGEXMATCH(A2,"Taiwan"),
B2-dataset!B11,REGEXMATCH(A2,"Russia"),
B2-dataset!B15,REGEXMATCH(A2,"Greece"),
B2-dataset!B19,REGEXMATCH(A2,"South Africa"),
B2-dataset!B23,REGEXMATCH(A2,"UAE"),
B2-dataset!B27,REGEXMATCH(A2,"Albania"),
B2-dataset!B31,REGEXMATCH(A2,"India"),
B2-dataset!B35,REGEXMATCH(A2,"South Korea"),
B2-dataset!B39,REGEXMATCH(A2,"Turkey"),
B2-dataset!B43)
The idea is that B2 (currently as =date(dd/mm/yyyy) has a deadline date. C2 (in which the formula houses) should show the date when everything should be delivered.
Currently the outcome is a number, not a date. I've tried IF statement, which delivers a date but I can only add 3 arguments. Can someone help me?
Kind regards
if your current output is number like 40000+ it's ok and it's just formatting issue. either you will format it internally or use a formula.
try:
=TEXT(IFS(regexmatch(A2,"Malaysia"),
B2-dataset!B3,REGEXMATCH(A2,"Saudi Arabia"),
B2-dataset!B7,REGEXMATCH(A2,"Taiwan"),
B2-dataset!B11,REGEXMATCH(A2,"Russia"),
B2-dataset!B15,REGEXMATCH(A2,"Greece"),
B2-dataset!B19,REGEXMATCH(A2,"South Africa"),
B2-dataset!B23,REGEXMATCH(A2,"UAE"),
B2-dataset!B27,REGEXMATCH(A2,"Albania"),
B2-dataset!B31,REGEXMATCH(A2,"India"),
B2-dataset!B35,REGEXMATCH(A2,"South Korea"),
B2-dataset!B39,REGEXMATCH(A2,"Turkey"),
B2-dataset!B43), "dd/mm/yyyy")
Using Binance Futures API I am trying to get a proper form of my position regarding cryptocurrencies.
Using the code
from binance_f import RequestClient
request_client = RequestClient(api_key= my_key, secret_key=my_secet_key)
result = request_client.get_position()
I get the following result
[{"symbol":"BTCUSDT","positionAmt":"0.000","entryPrice":"0.00000","markPrice":"5455.13008723","unRealizedProfit":"0.00000000","liquidationPrice":"0","leverage":"20","maxNotionalValue":"5000000","marginType":"cross","isolatedMargin":"0.00000000","isAutoAddMargin":"false"}]
The type command indicates it is a list, however adding at the end of the code print(result) yields:
[<binance_f.model.position.Position object at 0x1135cb670>]
Which is baffling because it seems not to be the list (in fact, debugging it indicates object of type Position). Using PrintMix.print_data(result) yields:
data number 0 :
entryPrice:0.0
isAutoAddMargin:True
isolatedMargin:0.0
json_parse:<function Position.json_parse at 0x1165af820>
leverage:20.0
liquidationPrice:0.0
marginType:cross
markPrice:5442.28502271
maxNotionalValue:5000000.0
positionAmt:0.0
symbol:BTCUSDT
unrealizedProfit:0.0
Now it seems like a JSON format... But it is a list. I am confused - any ideas how I can convert result to a proper DataFrame? So that columns are Symbol, PositionAmt, entryPrice, etc.
Thanks!
Your main question remains as you wrote on the header you should not be confused. In your case you have a list of Position object, you can see the structure of Position in the GitHub of this library
Anyway to answer the question please use the following:
df = pd.DataFrame([t.__dict__ for t in result])
For more options and information please read the great answers on this question
Good Luck!
you can use that
df = pd.DataFrame([t.__dict__ for t in result])
klines=df.values.tolist()
open = [float(entry[1]) for entry in klines]
high = [float(entry[2]) for entry in klines]
low = [float(entry[3]) for entry in klines]
close = [float(entry[4]) for entry in klines]
I have a column in data frame which ex df:
A
0 Good to 1. Good communication EI : tathagata.kar#ae.com
1 SAP ECC Project System EI: ram.vaddadi#ae.com
2 EI : ravikumar.swarna Role:SSE Minimum Skill
I have a list of of strings
ls=['tathagata.kar#ae.com','a.kar#ae.com']
Now if i want to filter out
for i in range(len(ls)):
df1=df[df['A'].str.contains(ls[i])
if len(df1.columns!=0):
print ls[i]
I get the output
tathagata.kar#ae.com
a.kar#ae.com
But I need only tathagata.kar#ae.com
How Can It be achieved?
As you can see I've tried str.contains But I need something for extact match
You could simply use ==
string_a == string_b
It should return True if the two strings are equal. But this does not solve your issue.
Edit 2: You should use len(df1.index) instead of len(df1.columns). Indeed, len(df1.columns) will give you the number of columns, and not the number of rows.
Edit 3: After reading your second post, I've understood your problem. The solution you propose could lead to some errors.
For instance, if you have:
ls=['tathagata.kar#ae.com','a.kar#ae.com', 'tathagata.kar#ae.co']
the first and the third element will match str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i])
And this is an unwanted behaviour.
You could add a check on the end of the string: str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i]+r'(?:\s|$)')
Like this:
for i in range(len(ls)):
df1 = df[df['A'].str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i]+r'(?:\s|$)')]
if len(df1.index != 0):
print (ls[i])
(Remove parenthesis in the "print" if you use python 2.7)
Thanks for the help. But seems like I found a solution that is working as of now.
Must use str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i])
This seems to solve the problem.
Although thanks to #IsaacDj for his help.
Why not just use:
df1 = df[df['A'].[str.match][1](ls[i])
It's the equivalent of regex match.
I want to display an output with Test = 0 using Fortran, I tried to use:
'WRITE(11,*) 'Test =' testdata'
Assuming 11 is correct and testdata is a parameter that is being calculated.
I wasn't able to get the output and there was an error.
Anyone have any idea why it is so?
Try inserting a comma and deleting the apostrophes:
WRITE(11,*) 'Test =', testdata
If you had reported what the error message you saw was I might have made this answer more apposite.
I have this code:
msgs = int(post['time_in_weeks'])
for i in range(msgs):
tip_msg = Tip.objects.get(week_number=i)
it always results in an error saying that no values could be found.
week_number is an integer field. When I input the value of i directly,
the query works.
When i print out the value of i I get the expected values.
Any input at all would be seriously appreciated.
Thanks.
The range function will give you a zero based list of numbers up to and excluding msgs. I guess there is no Tip with week_number=0.
Also, to limit the number of queries you could do this:
for tip in Tip.objects.filter(week_number__lt=msgs):
#do something
or, if you want specific weeks:
weeks=(1,3,5)
for tip in Tip.objects.filter(week_number__in=weeks):
#do something