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Python: merge columns with same name, keeping minimum value

I have a big matrix, like this:
df:
A A A B B ... (column names)
A 2 4 5 9 2
A 6 8 7 6 4
A 5 2 6 4 5
B 3 4 1 3 4
B 4 5 3 1 4
.
.
(row names)
I would like to merge the columns with same name, and findig the minimum value. At the end I would like to have a matrix like this:
df_min:
A B ... (column names)
A 2 2
A 6 4
A 2 4
B 1 3
B 3 1
.
.
(row names)
My intentions, afterwards (outside of the question), is to merge the rows as well. Desired outcome:
df_min:
A B ... (column names)
A 2 2
B 1 1
.
.
(row names)
I tried this:
df_min= df.groupby('df.columns, axis=1').agg(np.min)
But it didn't work, it removed some rows (for example, removing entirely row A)... EDIT: Apparently, it worked fine but I had two columns with different names but whitespace at the end of the name. These methods reorder the columns, which confused me.
A snipped of the dataframe:

Simply groupby on the level=0 for each axis:
df.groupby(level=0, axis=1).min()
output:
A B
A 2 2
A 6 4
A 2 4
B 1 3
B 3 1
both axes:
df.groupby(level=0, axis=1).min().groupby(level=0).min()
output:
A B
A 2 2
B 1 1
Alternatively, use a single groupby trough a stack/unstack:
df.stack().groupby(level=[0,1]).min().unstack()
output:
A B
A 2 2
B 1 1

EDIT
numpy only based solution
I'm assuming that you have a list associating names to column indices, e.g. for the first code sample you provided something like
column_names = ['A', 'A', 'A', 'B', 'B']
and that your data type is single-precision floating point. In this scenario, you can do something like the following:
unique_column_names = list(dict.fromkeys(column_names)) # get unique column names preserving original order
df_min = np.empty((df.shape[0], len(unique_column_names), dtype=np.float32) # allocate output array
for i, column_name in enumerate(unique_column_names): # iterate over unique column names
column_indices = [id for id in range(df.shape[1]) if column_names[id] == column_name] # extract all column indices having the same name
tmp = df[:, column_indices] # extract columns named as column_name
df_min[:, i] = np.amin(tmp, axis=1)] # take min by row and save result
Then, if you want to repeat the process by row, assuming you have another list associating row indices and names named row_names
unique_row_names = list(dict.fromkeys(row_names)) # get unique row names preserving order
df_final = np.empty((len(unique_row_names), len(unique_column_names), dtype=np.float32) # allocate final output
for j, row_name in enumerate(unique_row_names): # iterate over unique row names
row_indices = [id for id in range(df.shape[0]) if row_names[id] == row_name] # extract rows having row_name
tmp = df_min[row_indices, :] # extract rows named as row_name from the column-reduced matrix
df_final[j, :] = np.amin(tmp, axis=0) # take min by column and save result
The column-name and row-name association list for the final output are unique_column_names and unique_row_names

Selecting first n columns and last n columns with pandas

I am trying to select the first 2 columns and the last 2 column from a data frame by index with pandas and save it on the same dataframe.
is there a way to do that in one step?

You can use the iloc function to get the columns, and then pass in the indexes.
df.iloc[:,[0,1,-1,-2]]

You are looking for iloc:
df = pd.DataFrame([[1,2,3,4,5], [2,3,4,5,6], [3,4,5,6,7]], columns=['a','b','c','d','e'])
df.iloc[:,:2] # Grabs all rows and first 2 columns
df.iloc[:,-2:] # Grabs all rows and last 2 columns
pd.concat([df.iloc[:,:2],df.iloc[:,-2:]],axis=1) # Puts them together row wise

df = pd.DataFrame([[1,2,3,4,5], [2,3,4,5,6], [3,4,5,6,7]], columns=['a','b','c','d','e'])
df[['a','b','d','e']]
result
a b d e
0 1 2 4 5
1 2 3 5 6
2 3 4 6 7

Pandas: Adding incremental numbers to the suffix of repeating values of a column that are grouped by the value of another column and ordered by index

I am trying to add an underscore and incremental numbers to any repeating values ordered by index and within a group that is defined by another column.
For example, I would like the repeating values in the Chemistry column to have underscores and incremental numbers ordered by index and grouped by the Cycle column.
df = pd.DataFrame([[1,1,1,1,1,1,2,2,2,2,2,2], ['NaOH', 'H20', 'MWS', 'H20', 'MWS', 'NaOh', 'NaOH', 'H20', 'MWS', 'H20', 'MWS', 'NaOh']]).transpose()
df.columns = ['Cycle', 'Chemistry']
df
Original Table
So the output will look like the table in the link below:
Desired output table

IIUC:
pandas.Series.str.cat and cumcount
df['Chemistry'] = df.Chemistry.str.cat(
df.groupby(['Cycle', 'Chemistry']).cumcount().add(1).astype(str),
sep='_'
)
df
Cycle Chemistry
0 1 NaOH_1
1 1 H20_1
2 1 MWS_1
3 1 H20_2
4 1 MWS_2
5 1 NaOh_1
6 2 NaOH_1
7 2 H20_1
8 2 MWS_1
9 2 H20_2
10 2 MWS_2
11 2 NaOH_2

How to keep only the top n% rows of each group of a pandas dataframe?

I have seen a variant of this question asked that keeps the top n rows of each group in a pandas dataframe and the solutions use n as an absolute number rather than a percentage here Pandas get topmost n records within each group. However, in my dataframe, each group has different numbers of rows in it and I want to keep the top n% rows of each group. How would I approach this problem?

You can construct a Boolean series of flags and filter before you groupby. First let's create an example dataframe and look at the number of row for each unique value in the first series:
np.random.seed(0)
df = pd.DataFrame(np.random.randint(0, 2, (10, 3)))
print(df[0].value_counts())
0 6
1 4
Name: 0, dtype: int64
Then define a fraction, e.g. 50% below, and construct a Boolean series for filtering:
n = 0.5
g = df.groupby(0)
flags = (g.cumcount() + 1) <= g[1].transform('size') * n
Then apply the condition, set the index as the first series and (if required) sort the index:
df = df.loc[flags].set_index(0).sort_index()
print(df)
1 2
0
0 1 1
0 1 1
0 1 0
1 1 1
1 1 0
As you can see, the resultant dataframe only has 3 0 indices and 2 1 indices, in each case half the number in the original dataframe.

Here is another option which builds on some of the answers in the post you mentioned
First of all here is a quick function to either round up or round down. If we want the top 30% of rows of a dataframe 8 rows long then we would try to take 2.4 rows. So we will need to either round up or down.
My preferred option is to round up. This is because, for eaxample, if we were to take 50% of the rows, but had one group which only had one row, we would still keep that one row. I kept this separate so that you can change the rounding as you wish
def round_func(x, up=True):
'''Function to round up or round down a float'''
if up:
return int(x+1)
else:
return int(x)
Next I make a dataframe to work with and set a parameter p to be the fraction of the rows from each group that we should keep. Everything follows and I have commented it so that hopefully you can follow.
import pandas as pd
df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4],'value':[1,2,3,1,2,3,4,1,1]})
p = 0.30 # top fraction to keep. Currently set to 80%
df_top = df.groupby('id').apply( # group by the ids
lambda x: x.reset_index()['value'].nlargest( # in each group take the top rows by column 'value'
round_func(x.count().max()*p))) # calculate how many to keep from each group
df_top = df_top.reset_index().drop('level_1', axis=1) # make the dataframe nice again
df looked like this
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
df_top looks like this
id value
0 1 3
1 2 4
2 2 3
3 3 1
4 4 1

Avoiding looping through pandas dataframe for feature generation

I'm working on some code that generates features from a dataframe and adds these features as columns to the dataframe.
The trouble is I'm working with a time series so that for any given tuple, I need (let's say) 5 of the previous tuples to generate the corresponding feature for that tuple.
lookback_period = 5
df['feature1'] = np.zeros(len(df)) # preallocate
for index, row in df.iterrows():
if index < lookback_period:
continue
slice = df[index - lookback_period:index]
some_int = SomeFxn(slice)
row['feature1'] = some_int
Is there a way to execute this code without explicitly looping through every row and then slicing?
One way is to create several lagged columns using df['column_name'].shift() such that all the necessary information is contained in each row, but this quickly gets intractable for my computer's memory since the dataset is large (millions of rows).

I don't have enough reputation to comment so will just post it here.
Can't you use apply for your dataframe e.g.
df['feature1'] = df.apply(someRowFunction, axis=1)
where someRowFunction will accept the full row and you can perform whatever row based slice and logic you want to do.
--- updated ---
As we do not have much information about the dataframe and the required/expected output, I just based the answer on the information from the comments
Let's define a function that will take a DataFrame slice (based on current row index and lookback) and the row and will return sum of the first column of the slice and value of the current row.
def someRowFunction (slice, row):
if slice.shape[0] == 0:
return 0
return slice[slice.columns[0]].sum() + row.b
d={'a':[1,2,3,4,5,6,7,8,9,0],'b':[0,9,8,7,6,5,4,3,2,1]}
df=pd.DataFrame(data=d)
lookback = 5
df['c'] = df.apply(lambda current_row: someRowFunction(df[current_row.name -lookback:current_row.name],current_row),axis=1)
we can get row index from apply using its name attribute and as such we can retrieve the required slice. Above will result to the following
print(df)
a b c
0 1 0 0
1 2 9 0
2 3 8 0
3 4 7 0
4 5 6 0
5 6 5 20
6 7 4 24
7 8 3 28
8 9 2 32
9 0 1 36