Combining data from two dataframe columns into one column - python-2.7

I have time series data in two separate DataFrame columns which refer to the same parameter but are of differing lengths.
On dates where data only exist in one column, I'd like this value to be placed in my new column. On dates where there are entries for both columns, I'd like to have the mean value. (I'd like to join using the index, which is a datetime value)
Could somebody suggest a way that I could combine my two columns? Thanks.
Edit2: I written some code which should merge the data from both of my column, but I get a KeyError when I try to set the new values using my index generated from rows where my first df has values but my second df doesn't. Here's the code:
def merge_func(df):
null_index = df[(df['DOC_mg/L'].isnull() == False) & (df['TOC_mg/L'].isnull() == True)].index
df['TOC_mg/L'][null_index] = df[null_index]['DOC_mg/L']
notnull_index = df[(df['DOC_mg/L'].isnull() == True) & (df['TOC_mg/L'].isnull() == False)].index
df['DOC_mg/L'][notnull_index] = df[notnull_index]['TOC_mg/L']
df.insert(len(df.columns), 'Mean_mg/L', 0.0)
df['Mean_mg/L'] = (df['DOC_mg/L'] + df['TOC_mg/L']) / 2
return df
merge_func(sve)
And here's the error:
KeyError: "['2004-01-14T01:00:00.000000000+0100' '2004-03-04T01:00:00.000000000+0100'\n '2004-03-30T02:00:00.000000000+0200' '2004-04-12T02:00:00.000000000+0200'\n '2004-04-15T02:00:00.000000000+0200' '2004-04-17T02:00:00.000000000+0200'\n '2004-04-19T02:00:00.000000000+0200' '2004-04-20T02:00:00.000000000+0200'\n '2004-04-22T02:00:00.000000000+0200' '2004-04-26T02:00:00.000000000+0200'\n '2004-04-28T02:00:00.000000000+0200' '2004-04-30T02:00:00.000000000+0200'\n '2004-05-05T02:00:00.000000000+0200' '2004-05-07T02:00:00.000000000+0200'\n '2004-05-10T02:00:00.000000000+0200' '2004-05-13T02:00:00.000000000+0200'\n '2004-05-17T02:00:00.000000000+0200' '2004-05-20T02:00:00.000000000+0200'\n '2004-05-24T02:00:00.000000000+0200' '2004-05-28T02:00:00.000000000+0200'\n '2004-06-04T02:00:00.000000000+0200' '2004-06-10T02:00:00.000000000+0200'\n '2004-08-27T02:00:00.000000000+0200' '2004-10-06T02:00:00.000000000+0200'\n '2004-11-02T01:00:00.000000000+0100' '2004-12-08T01:00:00.000000000+0100'\n '2011-02-21T01:00:00.000000000+0100' '2011-03-21T01:00:00.000000000+0100'\n '2011-04-04T02:00:00.000000000+0200' '2011-04-11T02:00:00.000000000+0200'\n '2011-04-14T02:00:00.000000000+0200' '2011-04-18T02:00:00.000000000+0200'\n '2011-04-21T02:00:00.000000000+0200' '2011-04-25T02:00:00.000000000+0200'\n '2011-05-02T02:00:00.000000000+0200' '2011-05-09T02:00:00.000000000+0200'\n '2011-05-23T02:00:00.000000000+0200' '2011-06-07T02:00:00.000000000+0200'\n '2011-06-21T02:00:00.000000000+0200' '2011-07-04T02:00:00.000000000+0200'\n '2011-07-18T02:00:00.000000000+0200' '2011-08-31T02:00:00.000000000+0200'\n '2011-09-13T02:00:00.000000000+0200' '2011-09-28T02:00:00.000000000+0200'\n '2011-10-10T02:00:00.000000000+0200' '2011-10-25T02:00:00.000000000+0200'\n '2011-11-08T01:00:00.000000000+0100' '2011-11-28T01:00:00.000000000+0100'\n '2011-12-20T01:00:00.000000000+0100' '2012-01-19T01:00:00.000000000+0100'\n '2012-02-14T01:00:00.000000000+0100' '2012-03-13T01:00:00.000000000+0100'\n '2012-03-27T02:00:00.000000000+0200' '2012-04-02T02:00:00.000000000+0200'\n '2012-04-10T02:00:00.000000000+0200' '2012-04-17T02:00:00.000000000+0200'\n '2012-04-26T02:00:00.000000000+0200' '2012-04-30T02:00:00.000000000+0200'\n '2012-05-03T02:00:00.000000000+0200' '2012-05-07T02:00:00.000000000+0200'\n '2012-05-10T02:00:00.000000000+0200' '2012-05-14T02:00:00.000000000+0200'\n '2012-05-22T02:00:00.000000000+0200' '2012-06-05T02:00:00.000000000+0200'\n '2012-06-19T02:00:00.000000000+0200' '2012-07-03T02:00:00.000000000+0200'\n '2012-07-17T02:00:00.000000000+0200' '2012-07-31T02:00:00.000000000+0200'\n '2012-08-14T02:00:00.000000000+0200' '2012-08-28T02:00:00.000000000+0200'\n '2012-09-11T02:00:00.000000000+0200' '2012-09-25T02:00:00.000000000+0200'\n '2012-10-10T02:00:00.000000000+0200' '2012-10-24T02:00:00.000000000+0200'\n '2012-11-21T01:00:00.000000000+0100' '2012-12-18T01:00:00.000000000+0100'] not in index"

You are close, but you actually don't need to iterate over the rows when using the isnull() functions. by default
df[(df['DOC_mg/L'].isnull() == False) & (df['TOC_mg/L'].isnull() == True)].index
Will return just the index of the rows where DOC_mg/L is not null and TOC_mg/L is null.
Now you can do something like this to set the values for TOC_mg/L:
null_index = df[(df['DOC_mg/L'].isnull() == False) & \
(df['TOC_mg/L'].isnull() == True)].index
df['TOC_mg/L'][null_index] = df['DOC_mg/L'][null_index] # EDIT To switch the index position.
This will use the index of the rows where TOC_mg/L is null and DOC_mg/L is not null, and set the values for TOC_mg/L to the those found in DOC_mg/L in the same rows.
Note: This is not the accepted way for setting values using an index, but it is how I've been doing it for some time. Just make sure that when setting values, the left side of the equation is df['col_name'][index]. If col_name and index are switched you will set the values to a copy which is never set back to the original.
Now to set the mean, you can create a new column, we'll call this Mean_mg/L and set the value = 0.0. Then set this new column to the mean of both columns:
# Insert a new col at the end of the dataframe columns name 'Mean_mg/L'
# with default value 0.0
df.insert(len(df.columns), 'Mean_mg/L', 0.0)
# Set this columns value to the average of DOC_mg/L and TOC_mg/L
df['Mean_mg/L'] = (df['DOC_mg/L'] + df['TOC_mg/L']) / 2
In the columns where we filled null values with the corresponding column value, the average will be the same as the values.

Related

how to solve concatenate issue with.cell()? row = row work, column = column gives error

I am looping through an excel sheet, looking for a specific name. When found, I print the position of the cell and the value.
I would like to find the position and value of a neighbouring cell, however I can't get .cell() to work by adding 2, indicating I would like the cell 2 columns away in the same row.
row= row works, but column= column gives error, and column + 2 gives error. Maybe this is due to me listing columns as 'ABCDEFGHIJ' earlier in my code? (For full code, see below)
print 'Cell position {} has value {}'.format(cell_name, currentSheet[cell_name].value)
print 'Cell position next door TEST {}'.format(currentSheet.cell(row=row, column=column +2))
Full code:
file = openpyxl.load_workbook('test6.xlsx', read_only = True)
allSheetNames = file.sheetnames
#print("All sheet names {}" .format(file.sheetnames))
for sheet in allSheetNames:
print('Current sheet name is {}'.format(sheet))
currentSheet = file[sheet]
for row in range(1, currentSheet.max_row + 1):
#print row
for column in 'ABCDEFGHIJ':
cell_name = '{}{}'.format(column,row)
if currentSheet[cell_name].value == 'sign_name':
print 'Cell position {} has value {}'.format(cell_name, currentSheet[cell_name].value)
print 'Cell position TEST {}'.format(currentSheet.cell(row=row, column=column +2))
I get this output:
Current sheet name is Sheet1
Current sheet name is Sheet2
Cell position D5 has value sign_name
and:
TypeError: cannot concatenate 'str' and 'int' objects
I get the same error if I try "column = column" as "column = column +2".
Why does row=row work, but column=column dosen't? And how to find the cell name of the cell to the right of my resulting D5 cell?
The reason row=row works and column=column doesn't is because your column value is a string (letter from A to J) while the column argument of a cell is expecting an int (A would be 1, B would be 2, Z would be 26, etc.)
There are a few changes I would make in order to more effectively iterate through the cells and find a neighbor. Firstly, OpenPyXl offers sheet.iter_rows(), which given no arguments, will provide a generator of all rows that are used in the sheet. So you can iterate with
for row in currentSheet.iter_rows():
for cell in row:
because each row is a generator of cells in that row.
Then in this new nested for loop, you can get the current column index with cell.column (D would give 4) and the cell to the right (increment by one column) would be currentSheet.cell(row=row, column=cell.column+1)
Note the difference between the two cell's: currentSheet.cell() is a request for a specific cell while cell.column+1 is the column index of the current cell incremented by 1.
Relevant OpenPyXl documentation:
https://openpyxl.readthedocs.io/en/stable/api/openpyxl.cell.cell.html
https://openpyxl.readthedocs.io/en/stable/api/openpyxl.worksheet.worksheet.html

IF Formula in Power BI DAX

I would like to do a nested if statement within powerbi, I need to have multiple if statements in one column with some returning - value depending on the if statement.
I have tried to below however theyre all coming back as false.
Calculated value = if('TableName'[ColumnName1] = "exp1" && 'TableName'[columnName2] = "exp2", 'TableName'[value]|| if('TableName'[ColumnName1] = "exp1" && 'TableName'[ColumnName2] = "exp3", - 'TableName'[value],""))
In the first if you don't have any output to the false result. I Add an extra empty value in your calculated value as below only to exemplify what you need to add:
Calculated value = if('TableName'[ColumnName1] = "exp1" && 'TableName'[columnName2] = "exp2", 'TableName'[value]|| if('TableName'[ColumnName1] = "exp1" && 'TableName'[ColumnName2] = "exp3", - 'TableName'[value],""),"")
Check the example and tell something about this, please

Power Query conditional sumif

I need to add a column that Sums the value column of all columns that have a common id. However, any id = null is not summed, but equal to the value column.
The above example should result in:
TopPaymendId JournalLineNetAmount TopAmount
fcbcd407-ca26-4ea0-839a-c39767d05403 -3623.98 -7061.23
fcbcd407-ca26-4ea0-839a-c39767d05403 -3437.25 -7061.23
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This code should work:
let
Source = Excel.CurrentWorkbook(){[Name="Data"]}[Content],
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How to compare values in pandas between two different columns?

My Table:
A Country Code1 Code2
626349 US 640AD1237 407223
702747 NaN IO1062123 407255
824316 US NaN NaN
712947 US 00220221 870262123
278147 Canada 721AC31234 109123
278144 Canada NaN 7214234321
278142 Canada 72142QW134 109123AS12
Here in the above table I need to check country and code.
I want a 5th column with correct or wrong, pseudocode:
If 'Country' == 'US' and (length(Code1) OR length(Code2) == 9):
Add values to 5th column as correct.
else:
Add values to 5th column as incorrect.
If 'Country' == 'Canada' and (length(Code1) OR length(Code2) == 10):
Add values to 5th column as correct.
else:
Add values to 5th column as incorrect.
if no values are there either in Country or Code Column than insufficient information.
I am not able to understand how should I do this in pandas. Please help. Thanks.
I tried to first find the length of rows of Code1 and Code2 and store it in different df but after that I am not able to Compare the different set of data as what I need to do.
Len1 = df.Code1.map(len)
Len2 = df.Code2.map(len)
LengthCode = pd.DataFrame({'Len_Code1': Len1,'Len_Code2': Len2})
Please tell me the better way of how to do this in single dataframe if possible.
I tried this
df[(df.Country == 'US') & ((df.Code1.str.len() == 9)|(df.Code2.str.len() == 9))|(df.Country == 'Canada') & ((df.Code1.str.len() == 10)|(df.Code2.str.len() == 10))]
But it is getting long and I will not be able to write for many countries.
This will give you a 'is_correct' boolean column:
code_lengths = {'US':9, 'Canada':10}
df['correct_code_length'] = df.Country.replace(code_lengths)
df['is_correct'] = (df.Code1.apply(lambda x: len(str(x))) == df.correct_code_length) | (df.Code2.apply(lambda x: len(str(x))) == df.correct_code_length)
You will need to populate the code_lengths dictionary with more countries as necessary.

KeyError: Not in index, using a keys generated from a Pandas dataframe on itself

I have two columns in a Pandas DataFrame that has datetime as its index. The two column contain data measuring the same parameter but neither column is complete (some row have no data at all, some rows have data in both column and other data on in column 'a' or 'b').
I've written the following code to find gaps in columns, generate a list of indices of dates where these gaps appear and use this list to find and replace missing data. However I get a KeyError: Not in index on line 3, which I don't understand because the keys I'm using to index came from the DataFrame itself. Could somebody explain why this is happening and what I can do to fix it? Here's the code:
def merge_func(df):
null_index = df[(df['DOC_mg/L'].isnull() == False) & (df['TOC_mg/L'].isnull() == True)].index
df['TOC_mg/L'][null_index] = df[null_index]['DOC_mg/L']
notnull_index = df[(df['DOC_mg/L'].isnull() == True) & (df['TOC_mg/L'].isnull() == False)].index
df['DOC_mg/L'][notnull_index] = df[notnull_index]['TOC_mg/L']
df.insert(len(df.columns), 'Mean_mg/L', 0.0)
df['Mean_mg/L'] = (df['DOC_mg/L'] + df['TOC_mg/L']) / 2
return df
merge_func(sve)
Whenever you are considering performing assignment then you should use .loc:
df.loc[null_index,'TOC_mg/L']=df['DOC_mg/L']
The error in your original code is the ordering of the subscript values for the index lookup:
df['TOC_mg/L'][null_index] = df[null_index]['DOC_mg/L']
will produce an index error, I get the error on a toy dataset: IndexError: indices are out-of-bounds
If you changed the order to this it would probably work:
df['TOC_mg/L'][null_index] = df['DOC_mg/L'][null_index]
However, this is chained assignment and should be avoided, see the online docs
So you should use loc:
df.loc[null_index,'TOC_mg/L']=df['DOC_mg/L']
df.loc[notnull_index, 'DOC_mg/L'] = df['TOC_mg/L']
note that it is not necessary to use the same index for the rhs as it will align correctly