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Splitting a non delimited column and create an additional column to count which number value

I have a problem in which I want to take Table 1 and turn it into Table 2 using Python.
Does anybody have any ideas? I've tried to split the Value column from table 1 but run into issues in that each value is a different length, hence I can't always define how much to split it.
Equally I have not been able to think through how to create a new column that counts the position that value was in the string.
Table 1, before:
ID
Value
1
000000S
2
000FY
Table 2, after:
ID
Position
Value
1
1
0
1
2
0
1
3
0
1
4
0
1
5
0
1
6
0
1
7
S
2
1
0
2
2
0
2
3
0
2
4
F
2
5
Y

You can split the string to individual characters and explode:
out = (df
.assign(Value=df['Value'].apply(list))
.explode('Value')
)
output:
ID Value
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 S
1 2 0
1 2 0
1 2 0
1 2 F
1 2 Y

Given:
ID Value
0 1 000000S
1 2 000FY
Doing:
df.Value = df.Value.apply(list)
df = df.explode('Value')
df['Position'] = df.groupby('ID').cumcount() + 1
Output:
ID Value Position
0 1 0 1
0 1 0 2
0 1 0 3
0 1 0 4
0 1 0 5
0 1 0 6
0 1 S 7
1 2 0 1
1 2 0 2
1 2 0 3
1 2 F 4
1 2 Y 5

Sum specific number of columns for each row with Pandas

I have the dollowing dataframe:
name code 1 2 3 4 5 6 7 .........155 days
0 Lari EH214 0 5 2 1 0 0 0 0 3
1 Suzi FK362 0 0 0 0 2 3 0 0 108
2 Jil LM121 0 0 4 2 1 0 0 0 5
...
I want to sum the column between column 1 to column with the number that appears on "days" , for example,
for row 1, I will sum 3 days-> 0+5+2
For row 2 108 days,
for row 3 5 days->0+4+2+1+0
How can I do something like this?
Looking for method.

For vectorized solution filter rows by positions first and get mask by compare days in numpy boroadasting, if not match replace 0 in DataFrame.where and last sum:
df1 = df.iloc[:, 2:-1]
m = df1.columns.astype(int).to_numpy() <= df['days'].to_numpy()[:, None]
df['sum'] = df1.where(m, 0).sum(axis=1)
print (df)
name code 1 2 3 4 5 6 7 155 days sum
0 Lari EH214 0 5 2 1 0 0 0 0 3 7
1 Suzi FK362 0 0 0 0 2 3 0 0 108 5
2 Jil LM121 0 0 4 2 1 0 0 0 5 7

IIUC, use:
df['sum'] = df.apply(lambda r: r.loc[1: r['days']].sum(), axis=1)
or, if the column names are strings:
df['sum'] = df.apply(lambda r: r.loc['1': str(r['days'])].sum(), axis=1)
output:
name code 1 2 3 4 5 6 7 155 days sum
0 Lari EH214 0 5 2 1 0 0 0 0 3 7
1 Suzi FK362 0 0 0 0 2 3 0 0 108 5
2 Jil LM121 0 0 4 2 1 0 0 0 5 7

Cumulative count in a pandas df

I am trying to export a cumulative count based off two columns in a pandas df.
An example is the df below. I'm trying to export a count based off Value and Count. So when the count increase I want attribute that to the adjacent value
import pandas as pd
d = ({
'Value' : ['A','A','B','C','D','A','B','A'],
'Count' : [0,1,1,2,3,3,4,5],
})
df = pd.DataFrame(d)
I have used this:
for val in ['A','B','C','D']:
cond = df.Value.eq(val) & df.Count.eq(int)
df.loc[cond, 'Count_' + val] = cond[cond].cumsum()
If I alter int to a specific number it will return the count. But I need this to read any number as the Count column keeps increasing.
My intended output is:
Value Count A_Count B_Count C_Count D_Count
0 A 0 0 0 0 0
1 A 1 1 0 0 0
2 B 1 1 0 0 0
3 C 2 1 0 1 0
4 D 3 1 0 1 1
5 A 3 1 0 1 1
6 B 4 1 1 1 1
7 A 5 2 1 1 1
So the count increase on the second row so 1 to Value A. Count increases again on row 4 and it's the first time for Value C so 1. Same again for rows 5 and 7. The count increases on row 8 so A becomes 2.

You could use str.get_dummies and diff and cumsum
In [262]: df['Value'].str.get_dummies().multiply(df['Count'].diff().gt(0), axis=0).cumsum()
Out[262]:
A B C D
0 0 0 0 0
1 1 0 0 0
2 1 0 0 0
3 1 0 1 0
4 1 0 1 1
5 1 0 1 1
6 1 1 1 1
7 2 1 1 1
Which is
In [266]: df.join(df['Value'].str.get_dummies()
.multiply(df['Count'].diff().gt(0), axis=0)
.cumsum().add_suffix('_Count'))
Out[266]:
Value Count A_Count B_Count C_Count D_Count
0 A 0 0 0 0 0
1 A 1 1 0 0 0
2 B 1 1 0 0 0
3 C 2 1 0 1 0
4 D 3 1 0 1 1
5 A 3 1 0 1 1
6 B 4 1 1 1 1
7 A 5 2 1 1 1

Find first row with condition after each row satisfying another condition

in pandas I have the following data frame:
a b
0 0
1 1
2 1
0 0
1 0
2 1
Now I want to do the following:
Create a new column c, and for each row where a = 0 fill c with 1. Then c should be filled with 1s until the first row after each column fulfilling that, where b = 1 (and here im hanging), so the output should look like this:
a b c
0 0 1
1 1 1
2 1 0
0 0 1
1 0 1
2 1 1
Thanks!

It seems you need:
df['c'] = df.groupby(df.a.eq(0).cumsum())['b'].cumsum().le(1).astype(int)
print (df)
a b c
0 0 0 1
1 1 1 1
2 2 1 0
3 0 0 1
4 1 0 1
5 2 1 1
Detail:
print (df.a.eq(0).cumsum())
0 1
1 1
2 1
3 2
4 2
5 2
Name: a, dtype: int32

Apply a value to all instances of a number based on conditions

I have a df like this:
ID Number
1 0
1 0
1 1
2 0
2 0
3 1
3 1
3 0
I want to apply a 5 to any ids that have a 1 anywhere in the number column and a zero to those that don't. For example, if the number "1" appears anywhere in the Number column for ID 1, I want to place a 5 in the total column for every instance of that ID.
My desired output would look as such
ID Number Total
1 0 5
1 0 5
1 1 5
2 0 0
2 0 0
3 1 5
3 1 5
3 0 5
Trying to think of a way leverage applymap for this issue but not sure how to implement.

Use transform to add a column to your df as a result of a groupby on 'ID':
In [6]:
df['Total'] = df.groupby('ID').transform(lambda x: 5 if (x == 1).any() else 0)
df
Out[6]:
ID Number Total
0 1 0 5
1 1 0 5
2 1 1 5
3 2 0 0
4 2 0 0
5 3 1 5
6 3 1 5
7 3 0 5

You can use DataFrame.groupby() on ID column and then take max() of the Number column, and then make that into a dictionary and then use that to create the 'Total' column. Example -
grouped = df.groupby('ID')['Number'].max().to_dict()
df['Total'] = df.apply((lambda row:5 if grouped[row['ID']] else 0), axis=1)
Demo -
In [44]: df
Out[44]:
ID Number
0 1 0
1 1 0
2 1 1
3 2 0
4 2 0
5 3 1
6 3 1
7 3 0
In [56]: grouped = df.groupby('ID')['Number'].max().to_dict()
In [58]: df['Total'] = df.apply((lambda row:5 if grouped[row['ID']] else 0), axis=1)
In [59]: df
Out[59]:
ID Number Total
0 1 0 5
1 1 0 5
2 1 1 5
3 2 0 0
4 2 0 0
5 3 1 5
6 3 1 5
7 3 0 5