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Iterate over columns, find selection, create new column


By : Nicholas Leighton
Date : August 01 2020, 09:00 AM
help you fix your problem Well, it's definately not clean. A Python pro would definately have their concerns, but as I'm new to Python and Pandas I just gave it a shot using some Join, Regular Expressions and Split:
code :
import pandas as pd
df = pd.DataFrame({'color':['black','red','19sf','deep'],
                   'gender':['male','woman','black','1909sf'],
                   'model1':['A24','A1','Male','2202'],
                   'model2':['B2','B3','A22','blue'],
                   'extra_col1':['','','B1','woman'],
                   'extra_col2':['','','','A1'],
                   'extra_col3':['','','','B22']})
df = df[df.columns[0:]].apply(lambda x: '|'.join(x.dropna().astype(str)), axis=1)
df = df.replace(r'.*?((?:\w+\|){3}\w+(?=\|*$)).*', value = r'\1', regex = True)
df = df.str.split('|', expand = True)
df.columns = ['color','gender','model1','model2']
print(df)


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How to iterate over pairs of columns in a data table to create a new column in R


By : Qusay AH
Date : March 29 2020, 07:55 AM
Does that help I currently have data of complex numbers in the form: , To extract a column you can either use:
code :
dat$column
dat["column",] # Notice the "
dat[paste(...),]
dat$paste(...)

VBA - Macro to extend selection to end of current table column, then find and replace only within that selection


By : user7132892
Date : March 29 2020, 07:55 AM
To fix this issue I was able to create a range based on the selection.
Creates a selection from the current table cell to the end of the column:
code :
Selection.EndKey Unit:=wdColumn, Extend:=wdExtend
Dim myRange As Range
Set myRange = ActiveDocument.Range(Selection.Range.Start, Selection.Range.End)
With myRange.Find
  .Text = " "
  .Replacement.Text = ""
  .Wrap = wdFindStop
  .Execute Replace:=wdReplaceAll
End With

With myRange.Find
  .Text = "0"
  .Replacement.Text = "-"
  .Wrap = wdFindStop
  .Execute Replace:=wdReplaceAll
End With

myRange.Find.Execute Replace:=wdReplaceAll

Google spreadsheet Script - iterate on rows of active selection, but all columns


By : Klaus Laube
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I have read many similar questions but I still can't get my code to work and I believe this is because of a slight difference in what I want to achieve : , This code works properly for me
code :
  var sheet = SpreadsheetApp.getActiveSpreadsheet();
  var range = sheet.getActiveRange()
  var firstRow = range.getRow()
  var numRows = range.getNumRows();
  Logger.log("firstRow "+firstRow);
  Logger.log("numRows "+numRows);

  for (var i = 0; i < numRows; i++) {
    var absoluteRow = firstRow + i;
    Logger.log("absoluteRow "+absoluteRow); 
  }  
[16-09-01 13:44:44:762 EEST] firstRow 2
[16-09-01 13:44:44:763 EEST] numRows 5
[16-09-01 13:44:44:764 EEST] absoluteRow 2
[16-09-01 13:44:44:764 EEST] absoluteRow 3
[16-09-01 13:44:44:765 EEST] absoluteRow 4
[16-09-01 13:44:44:765 EEST] absoluteRow 5
[16-09-01 13:44:44:766 EEST] absoluteRow 6

Iterate over rows in a data frame create a new column then adding more columns based on the new column


By : Brent Sabo
Date : March 29 2020, 07:55 AM
hope this fix your issue I have a data frame as below: , IIUC using resample and interpolate, then we pivot the output
code :
s=df.set_index('Date').resample('1 H').interpolate()
s=pd.pivot_table(s,index=s.index.date,columns=s.groupby(s.index.date).cumcount(),values=s,aggfunc='mean')
s.columns=s.columns.droplevel(0)
s
Out[93]: 
               0      1      2      3   ...       20     21     22     23
2019-04-25  100.0  102.0  104.0  106.0  ...    140.0  142.0  144.0  146.0
2019-04-26  148.0  147.0  146.0  145.0  ...    128.0  127.0  126.0  125.0
2019-04-27  124.0    NaN    NaN    NaN  ...      NaN    NaN    NaN    NaN
[3 rows x 24 columns]

Iterate through columns and divide column value and creating new column with result in column


By : user2910874
Date : March 29 2020, 07:55 AM
Any of those help This line df['{}'.format(n)].iloc = df[s].iloc/5 is plain wrong:
s is not a column name but is a pair (column_name, column) so it cannot be used to index a dataframe. iloc is wrong too, and is anyway useless it is extremely dangerous to change something you are iterating, so in a for s in df.iteritems(): loop, you should never add columns to df
code :
cols = df.columns.tolist()
for j, s in enumerate(cols):  # ok cols is a plain list
    for i in range(5):
        df[str(i + 5*j)] = df[s]/5
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