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How to get some string of dataframe column?

I have dataframe like this.
print(df)
[ ID ... Control
0 PDF-1 ... NaN
1 PDF-3 ... NaN
2 PDF-4 ... NaN
I want to get only number of ID column. So the result will be.
1
3
4
How to get one of the strings of the dataframe column ?

How about just replace a common PDF- prefix?
df['ID'].str.replace('PDF-', '')

Could you please try following.
df['ID'].replace(regex=True,to_replace=r'([^\d])',value=r'')
One could refer documentation for df.replace
Basically using regex to remove everything apart from digits in column named ID where \d denotes digits and when we use [^\d] means apart form digits match everything.

Another possibility using Regex is:
df.ID.str.extract('(\d+)')
This avoids changing the original data just to extract the integers.
So for the following simple example:
import pandas as pd
df = pd.DataFrame({'ID':['PDF-1','PDF-2','PDF-3','PDF-4','PDF-5']})
print(df.ID.str.extract('(\d+)'))
print(df)
we get the following:
0
0 1
1 2
2 3
3 4
4 5
ID
0 PDF-1
1 PDF-2
2 PDF-3
3 PDF-4
4 PDF-5

Find "PDF-" ,and replace it with nothing
df['ID'] = df['ID'].str.replace('PDF-', '')
Then to print how you asked I'd convert the data frame to a string with no index.
print df['cleanID'].to_string(index=False)

Parse Out Last Sequence Of Numbers From Pandas Column to create new column

I have a dataframe with codes like the following and would like to create a new column that has the last sequence of numbers parse out.
array(['K9ADXXL2', 'K9ADXL2', 'K9ADXS2', 'IVERMAXSCM12', 'HPDMUDOGDRYL'])
So the new column would contain the following:
array([2,2,2,12,None])

Sample data
df:
codes
0 K9ADXXL2
1 K9ADXL2
2 K9ADXS2
3 IVERMAXSCM12
4 HPDMUDOGDRYL
Use str.extract gets digits at the end of string and passing to pd.to_numeric
pd.to_numeric(df.codes.str.extract(r'(\d+$)')[0], errors='coerce')
Out[11]:
0 2.0
1 2.0
2 2.0
3 12.0
4 NaN
Name: 0, dtype: float64
If you want get value as string of numbers, you may use str.extract or str.findall as follow
df.codes.str.findall(r'\d+$').str[0]
or
df.codes.str.extract(r'(\d+$)')[0]
Out[20]:
0 2
1 2
2 2
3 12
4 NaN
Name: codes, dtype: object

import re
import pandas as pd
def get_trailing_digits(s):
match = re.search("[0-9]+$",s)
return match.group(0) if match else None
original_column = pd.array(['K9ADXXL2', 'K9ADXL2', 'K9ADXS2', 'IVERMAXSCM12', 'HPDMUDOGDRYL'])
new_column = pd.array([get_trailing_digits(s) for s in original_column])
# ['2', '2', '2', '12', None]
0-9] means any digit
+ means one or more times
$means only at the end of the string
You can use the apply function of a series/data frame with get_trailing_digits as the function.
eg.
my_df["new column"] = my_df["old column"].apply(get_trailing_digits)

Pandas - slicing column values based on another column

How can I slice column values based on first & last character location indicators from two other columns?
Here is the code for a sample df:
import pandas as pd
d = {'W': ['abcde','abcde','abcde','abcde']}
df = pd.DataFrame(data=d)
df['First']=[0,0,0,0]
df['Last']=[1,2,3,5]
df['Slice']=['a','ab','abc','abcde']
print(df.head())
Code output:
Desired Output:

Just do it with for loop , you may worry about the speed , please check For loops with pandas - When should I care?
df['Slice']=[x[y:z]for x,y,z in zip(df.W,df.First,df.Last)]
df
Out[918]:
W First Last Slice
0 abcde 0 1 a
1 abcde 0 2 ab
2 abcde 0 3 abc
3 abcde 0 5 abcde

I am not sure if this will be faster, but a similar approach would be:
df['Slice'] = df.apply(lambda x: x[0][x[1]:x[2]],axis=1)
Briefly, you go through each row (axis=1) and apply a custom function. The function takes the row (stored as x), and slices the first element using the second and third elements as indices for the slicing (that's the lambda part). I will be happy to elaborate more if this isn't clear.

How to add all values across pandas columns when the names of these columns match those in a list?

list = ['abc', 'def_1', 'xyz_8']
An example row for a df below
abc_1 abc_99 def_1 def_2 xyz_8 xyz_1
2 1 1 2 2 3
I would like to scan and select only some of the df columns based on the list. The list element can be a substring of the column name. For example column abc_1 will be included since abc is a substring, but xyz_1 is not included since xyz_1 is not an element of the list, and none of the list element is a substring of xyz_1.
I want a df['sum'] = 6 (or 2+1+1+2) for that row.

filter / str.contains
You can use filter or str.contains, both of which supports regex:
L = ['abc', 'def_1', 'xyz_8']
# courtesy of #JonClements
df['result'] = df.filter(regex='|'.join(L)).sum(1)
# original
df['result'] = df.iloc[:, df.columns.str.contains('|'.join(L))].sum(1)
print(df)
abc_1 abc_99 def_1 def_2 xyz_8 xyz_1 result
0 2 1 1 2 2 3 6

Python filling string column "forward" and groupby attaching groupby result to dataframe

I have a dataframe looking generated by:
df = pd.DataFrame([[100, ' tes t ', 3], [100, np.nan, 2], [101, ' test1', 3 ], [101,' ', 4]])
It looks like
0 1 2
0 100 tes t 3
1 100 NaN 2
2 101 test1 3
3 101 4
I would like to a fill column 1 "forward" with test and test1. I believe one approach would be to work with replacing whitespace by np.nan, but it is difficult since the words contain whitespace as well. I could also groupby column 0 and then use the first element of each group to fill forward. Could you provide me with some code for both alternatives I do not get it coded?
Additionally, I would like to add a column that contains the group means that is
the final dataframe should look like this
0 1 2 3
0 100 tes t 3 2.5
1 100 tes t 2 2.5
2 101 test1 3 3.5
3 101 test1 4 3.5
Could you also please advice how to accomplish something like this?
Many thanks please let me know in case you need further information.

IIUC, you could use str.strip and then check if the stripped string is empty.
Then, perform groupby operations and filling the Nans by the method ffill and calculating the means using groupby.transform function as shown:
df[1] = df[1].str.strip().dropna().apply(lambda x: np.NaN if len(x) == 0 else x)
df[1] = df.groupby(0)[1].fillna(method='ffill')
df[3] = df.groupby(0)[2].transform(lambda x: x.mean())
df
Note: If you must forward fill NaN values with first element of that group, then you must do this:
df.groupby(0)[1].apply(lambda x: x.fillna(x.iloc[0]))
Breakup of steps:
Since we want to apply the function only on strings, we drop all the NaN values present before, else we would be getting the TypeError due to both floats and string elements present in the column and complains of float having no method as len.
df[1].str.strip().dropna()
0 tes t # operates only on indices where strings are present(empty strings included)
2 test1
3
Name: 1, dtype: object
The reindexing part isn't a necessary step as it only computes on the indices where strings are present.
Also, the reset_index(drop=True) part was indeed unwanted as the groupby object returns a series after fillna which could be assigned back to column 1.