I've got a tricky situation - tricky for me since I'm really new to python. I've got a dataframe in pandas and I need to logic my way through building a new column that will be used later in a data match from a difference source. Basically, the picture tells what I can't figure out.
For any of the LOW labels I need to retrieve their MID_LEVEL label and copy it to a new column. The DESIRED OUTPUT column is what I need to create.
You can see that the LABEL_PATH is formatted in a way that I can use the first 9 digits as a "lookup" to find the corresponding LABEL, but I can't figure out how to achieve that. As an example, for any row that the LABEL_PATH starts with "0.02.0004" the desired output needs to be "MID_LEVEL1".
This dataset has around 25k rows, so wanted to avoid row iteration as well.
Any help would be greatly appreciated!
Chosing a similar example as you did:
df = pd.DataFrame({"a":["1","1.1","1.1.1","1.1.2","2"],"b":range(5)})
df["c"] = np.nan
mask = df.a.apply(lambda x: len(x.split(".")) < 3)
df.loc[mask,"c"] = df.b[mask]
df.c.fillna(method="ffill", inplace=True)
Most of the magic takes place in the line where mask is defined, but it's not that difficult: if the value in a gets split into less than 3 parts (i.e., has at most one dot), mark it as True, otherwise not.
Use that mask to copy over the values, and then fill unspecified values with valid values from above.
I am using this data for comparison :
test_dict = {"label_path": [1, 2, 3, 4, 5, 6], "label": ["low1", "low2", "mid1", "mid2", "high1", "high2"], "desired_output": ["mid1", "mid2", "mid1", "mid2", "high1", "high2"]}
df = pd.DataFrame(test_dict)
Which gives :
label_path label desired_output
0 1 low1 mid1
1 2 low2 mid2
2 3 mid1 mid1
3 4 mid2 mid2
4 5 high1 high1
5 6 high2 high2
With a bit ogf logic and a merge :
desired_label_df = df.drop_duplicates("desired_output", keep="last")
desired_label_df = desired_label_df[["label_path", "desired_output"]]
desired_label_df.columns = ["desired_label_path", "desired_output"]
df = df.merge(desired_label_df, on="desired_output", how="left")
Gives us :
label_path label desired_output desired_label_path
0 1 low1 mid1 3
1 2 low2 mid2 4
2 3 mid1 mid1 3
3 4 mid2 mid2 4
4 5 high1 high1 5
5 6 high2 high2 6
Edit: if you want to create the desired_output column, just do the following :
df["desired_output"] = df["label"].apply(lambda x: x.replace("low", "mid"))
Related
I have a dataframe with a column that contains tuples. I would like to create a new column that extracts the smallest tuple from the tuple column.
What I have tried so far
mydataframe['min_values'] = mydataframe['tuple_column'].apply(lambda x: min(x))
This above approach seems to work when I have at least 2 tuples, but it fails when I only have one tuple e.g. 5 in the example below. Could you guys please suggest a method that would help me accomplish this task in a better manner?
Example and desired result
Tuple Column
New Column
(1,2,3,5)
1
(10,11)
10
(5)
5
Thanks
(5) is not a tuple, this is 5. Use numpy.min that handles scalar values as input:
import numpy as np
df['New Column'] = df['Tuple Column'].apply(np.min)
Output:
Tuple Column New Column
0 (1, 2, 3, 5) 1
1 (10, 11) 10
2 5 5
Here is a way using map()
df['Tuple Column'].map(lambda x: min(x) if isinstance(x,tuple) else x)
Output:
0 1
1 3
2 5
df1.applymap(lambda x:pd.Series(eval(x)).min())
Output:
0 1
1 3
2 5
I have multiple Excel spreadsheets containing the same types of data but they are not in the same order. For example, if file 1 has the results of measurements A, B, C and D from River X printed in columns 1, 2, 3 and 4, respectively but file 2 has the same measurements taken for a different river, River Y, printed in columns 6, 7, 8, and 9 respectively, is there a way to use pandas to reorganise one dataframe to match the layout of another dataframe (i.e. make it so that Sheet2 has the measurements for River Y printed in columns 1, 2, 3 and 4)? Sometimes the data is presented horizontally, not vertically as described above, too. If I have the same measurements for, say, 400 different rivers on 400 separate sheets, but the presentation/layout of data is erratic with regards to each individual file, it would be useful to be able to put a single order on every spreadsheet without having to manually shift columns on Excel.
Is there a way to use pandas to reorganise one dataframe to match the layout of another dataframe?
You can get a list of columns from one of your dataframes and then sort that. Next you can use the sorted order to reorder your remaining dataframes. I've created an example below:
import pandas as pd
import numpy as np
# Create an example of your problem
root = 'River'
suffix = list('123')
cols_1 = [root + '_' + each_suffix for each_suffix in suffix]
cols_2 = [root + '_' + each_suffix for each_suffix in suffix[::]]
data = np.arange(9).reshape(3,3)
df_1 = pd.DataFrame(columns=cols_1, data=data)
df_2 = pd.DataFrame(columns=cols_2, data=data)
df_1
[out] River_1 River_2 River_3
0 0 1 2
1 3 4 5
2 6 7 8
df_2
[out] River_3 River_2 River_1
0 0 1 2
1 3 4 5
2 6 7 8
col_list = df_1.columns.to_list() # Get a list of column names use .sort() to sort in place or
sorted_col_list = sorted(col_list, reverse=False) # Use reverse True to invert the order
def rearrange_df_cols(df, target_order):
df = df[target_order]
print(df)
return df
rearrange_df_cols(df_1, sorted_col_list)
[out] River_1 River_2 River_3
0 0 1 2
1 3 4 5
2 6 7 8
rearrange_df_cols(df_2, sorted_col_list)
[out] River_1 River_2 River_3
0 2 1 0
1 5 4 3
2 8 7 6
You can write a function based on what's above and apply it to all of your file/sheets provided that all columns names exist (NB the must be written identically).
Sometimes the data is presented horizontally, not vertically as described above, too.
This would be better as a separate question. In principle you should check the dimension of your data e.g. df.shape and based of the shape you can either use df.transpose() and then your function to reorder the columns names or directly use your function to reorder the column names.
I've got a Dataframe like this:
df = pd.DataFrame(np.reshape(np.arange(0,9), (3,3)))
print(df)
0 1 2
0 0 1 2
1 3 4 5
2 6 7 8
I'd like to normalize two of the columns against a reference column. For example, if I chose df[0] as my reference column, then df[1] and df[2] would also have a mean of 3 and a standard deviation of 3.
What's the best way to do this?
You can shift and scale the values in each column by the mean and standard deviation of the reference column ref:
ref = 0
means = df.mean()
stds = df.std()
(df - means + means[ref]) / stds * stds[ref]
BACKGROUND: Large excel mapping file with about 100 columns and 200 rows converted to .csv. Then stored as dataframe. General format of df as below.
Starts with a named column (e.g. Sales) and following two columns need to be renamed. This pattern needs to be repeated for all columns in excel file.
Essentially: Link the subsequent 2 columns to the "parent" one preceding them.
Sales Unnamed: 2 Unnamed: 3 Validation Unnamed: 5 Unnamed: 6
0 Commented No comment Commented No comment
1 x x
2 x x
3 x x
APPROACH FOR SOLUTION: I assume it would be possible to begin with an index (e.g. index of Sales column 1 = x) and then rename the following two columns as (x+1) and (x+2).
Then take in the text for the next named column (e.g. Validation) and so on.
I know the rename() function for dataframes.
BUT, not sure how to apply the iteratively for changing column titles.
EXPECTED OUTPUT: Unnamed 2 & 3 changed to Sales_Commented and Sales_No_Comment, respectively.
Similarly Unnamed 5 & 6 change to Validation_Commented and Validation_No_Comment.
Again, repeated for all 100 columns of file.
EDIT: Due to the large number of cols in the file, creating a manual list to store column names is not a viable solution. I have already seen this elsewhere on SO. Also, the amount of columns and departments (Sales, Validation) changes in different excel files with the mapping. So a dynamic solution is required.
Sales Sales_Commented Sales_No_Comment Validation Validation_Commented Validation_No_Comment
0 Commented No comment Commented No comment
1 x x
2 x
3 x x x
As a python novice, I considered a possible approach for the solution using the limited knowledge I have, but not sure what this would look like as a workable code.
I would appreciate all help and guidance.
1.You need is to make a list with the column names that you would want.
2.Make it a dict with the old column names as the keys and new column name as the values.
3. Use df.rename(columns = your_dictionary).
import numpy as np
import pandas as pd
df = pd.read_excel("name of the excel file",sheet_name = "name of sheet")
print(df.head())
Output>>>
Sales Unnamed : 2 Unnamed : 3 Validation Unnamed : 5 Unnamed : 6 Unnamed :7
0 NaN Commented No comment NaN Comment No comment Extra
1 1.0 2 1 1.0 1 1 1
2 3.0 1 1 1.0 1 1 1
3 4.0 3 4 5.0 5 6 6
4 5.0 1 1 1.0 21 3 6
# get new names based on the values of a previous named column
new_column_names = []
counter = 0
for col_name in df.columns:
if (col_name[:7].strip()=="Unnamed"):
new_column_names.append(base_name+"_"+df.iloc[0,counter].replace(" ", "_"))
else:
base_name = col_name
new_column_names.append(base_name)
counter +=1
# convert to dict key pair
dictionary = dict(zip(df.columns.tolist(),new_column_names))
# rename columns
df = df.rename(columns=dictionary)
# drop first column
df = df.iloc[1:].reset_index(drop=True)
print(df.head())
Output>>
Sales Sales_Commented Sales_No_comment Validation Validation_Comment Validation_No_comment Validation_Extra
0 1.0 2 1 1.0 1 1 1
1 3.0 1 1 1.0 1 1 1
2 4.0 3 4 5.0 5 6 6
3 5.0 1 1 1.0 21 3 6
I am inputting multiple spreadsheets with multiple columns of data. For each spreadsheet, the maximum value of each column is found. Then, for each element in the column, the element is divided by the maximum value of that column. The output should be a value (between 0 and 1) for each element in the column in ascending order. This is appended to a list which should be added to the source spreadsheet as a column.
Currently, the nested loops are performing correctly apart from the final step, as far as I understand. Each column is added to the spreadsheet EXCEPT the values are for the final column of the source spreadsheet rather than values related to each individual column.
I have tried changing the indents to associate levels of the code with different parts (as I think this is the problem) and tried moving the appended column along in the dataframe, to no avail.
for i in distlist:
#listname = i[4:] + '_norm'
df2 = pd.read_excel(i,header=0,index_col=None, skip_blank_lines=True)
df3 = df2.dropna(axis=0, how='any')
cols = []
for column in df3:
cols.append(column)
for x in cols:
listname = x + ' norm'
maxval = df3[x].max()
print(maxval)
mylist = []
for j in df3[x]:
findNL = (j/maxval)
mylist.append(findNL)
df3[listname] = mylist
saveloc = 'E:/test/'
filename = i[:-18] + '_Normalised.xlsx'
df3.to_excel(saveloc+filename, index=False)
New columns are added to the output dataframe with bespoke headings relating to the field headers in the source spreadsheet and renamed according to (listname). The data in each one of these new columns is identical and relates to the final column in the spreadsheet. To me, it seems to be overwriting the values each time (as if looping through the entire spreadsheet, not outputting for each column), and adding it to the spreadsheet.
Any help would be much appreciated. I think it's something simple, but I haven't managed to work out what...
If I understand you correctly, you are overcomplicating things. You dont need a for loop for this. You can simplify your code:
# Make example dataframe, this is not provided
df = pd.DataFrame({'col1':[1, 2, 3, 4],
'col2':[5, 6, 7, 8]})
print(df)
col1 col2
0 1 5
1 2 6
2 3 7
3 4 8
Now we can use DataFrame.apply and use add_suffix to give the new columns _norm suffix and after that concat the columns to one final dataframe
df_conc = pd.concat([df, df.apply(lambda x: x/x.max()).add_suffix('_norm')],axis=1)
print(df_conc)
col1 col2 col1_norm col2_norm
0 1 5 0.25 0.625
1 2 6 0.50 0.750
2 3 7 0.75 0.875
3 4 8 1.00 1.000
Many thanks. I think I was just overcomplicating it. Incidentally, I think my code may do the same job, but because there is so little difference in the values, it wasn't notable.
Thanks for your help #Erfan