Utilizing Pandas functionality/making code more pythonic to rewrite an excel macro-CodePudding

so I am rewriting an excel macro, and while I have successfully recreated it; I want to redo my code because it is a bit simplistic and is more of a simpler/brute force approach to the solution. I realized I am not really utilizing pandas functionality to it's fullest extent (am new to pandas, not to python). I also didn't have the most time to develop this so I just went.

There are a solid 30 formulas I need to recreate and they are basically all if statements nested within this starting for loop (these aren't real variable names):

for row in range(len(dataframe)):
    if dataframe.iat[row, 8][12:-2] == "some string":
        dataframe.iat[row, 4] = "some other string"
        
    # convert float to int into a string
    elif dataframe.iat[row, 8][10:-4] == str(int(dataframe.iat[row, 13]))[2:]:
        dataframe.iat[row, 4] = "some string2"
    else:
        dataframe.iat[row, 4] = "some string3"

Example of this code in VBA form:

=IF(MID(I2,13,2)="some string","some other string",(IF(RIGHT(N2,2)=MID(I2,11,2),"some string2","some string3")))

One more example (trust me I know using this many 'or's is not really pythonic at all - it will be fixed):

    if dataframe.iat[row, 27] != 0 or dataframe.iat[row, 28] != 0 or dataframe.iat[row, 29] 
              != 0 or dataframe.iat[row, 25] != 0 or dataframe.iat[row, 30] != 0:
        dataframe.iat[row, 23] = "Yes"
    else:
        dataframe.iat[row, 23] = "No"

The biggest problem I am facing is when I have to access multiple columns and it is probably due to my lack of pandas experience.

CodePudding user response：

It is tricky to give really solid answers without example data, however, I think .loc can solve most of this.

Example:

dataframe.loc[(dataframe["Column 8's name"].str[12:-2] == "some string"), "Column 4s name"] = "some other string"
dataframe.loc[(dataframe["Column 8's name"].str[10:-4] == dataframe["Column 13's name"].map(int).map(str).str[2:), "Column 4s name"] = "some string2"

# then finally just fill 
dataframe["Column 4s name"] = dataframe["Column 4s name"].fillna("some string3")

.loc is a much faster method than iterating through the rows, as it uses vectorization. I have also included the method .map, which replaces the built in str() and int() method.