CODEWITHSUNDEEP
pythonpandas
Akshad Patil
Akshad Patil

Reputation: 47

search for a value in pandas dataframe in specific column

for the input such as name AARTIIND and ltp 1014 I want output as url corresponding to lower bound of the ltp,

in this case https: // kite.zerodha.com/chart/ext/tvc/NFO-OPT/AARTIIND21JUN1020CE/13192450

I am trying to write function to do this, I am stuck here, unable to think what should be the searching condition which will return correct url

def find_url(name,ltp,df):
    return(df.iloc[df.index == name, ???)


name        strike      url
AARTIIND    1000    https: // kite.zerodha.com/chart/ext/tvc/NFO-OPT/AARTIIND21JUN1000CE/13190658
AARTIIND    1020    https: // kite.zerodha.com/chart/ext/tvc/NFO-OPT/AARTIIND21JUN1020CE/13192450
AARTIIND    1040    https: // kite.zerodha.com/chart/ext/tvc/NFO-OPT/AARTIIND21JUN1040CE/13201410
AARTIIND    1060    https: // kite.zerodha.com/chart/ext/tvc/NFO-OPT/AARTIIND21JUN1060CE/13202434

how can I get the required result?

Upvotes: 1

Views: 126

Answers (1)

Corralien
Corralien

Reputation: 120391

To find the closest value, subtract ltp to strike and compute the absolute value. The lowest value is the closest.

Try this:

def find_url(name, ltp, df):
    df1 = df.loc[df["name"] == name]
    return df1.loc[df1["strike"].sub(ltp).abs().idxmin(), "url"]

url = find_url("AARTIIND", 1014, df)

>>> url
'https: // kite.zerodha.com/chart/ext/tvc/NFO-OPT/AARTIIND21JUN1020CE/13192450'

Edit:

>>> pd.DataFrame({'df["strike"]': df["strike"],
                  ".sub(ltp)": df["strike"].sub(ltp),
                  ".abs()": df["strike"].sub(ltp).abs()})

   df["strike"]  .sub(ltp)  .abs()
0          1000        -14      14
1          1020          6       6  # idxmin() return 1 (6 == the lowest value)
2          1040         26      26
3          1060         46      46

Upvotes: 1

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