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Appending New Row to Existing Dataframe using Pandas

I'm trying to using pandas to append a blank row based on the values in the first column. When the first six characters in the first column don't match, I want an empty row between them (effectively creating groups). Here is an example of what the output could look like:
002446
002447-01
002447-02
002448
This is what I was able to put together thus far.
readie=pd.read_csv('title.csv')
i=0
for row in readie:
readie.append(row)
i+=1
if readie['column title'][i][0:5]!=readie['column title'][i+1][0:5]:
readie.append([])
When running this code, I get the following error message:
TypeError: cannot concatenate object of type '<class 'str'>'; only Series and DataFrame objs are valid
I believe there are other ways to do this, but I would like to use pandas if at all possible.

I'm using the approach from this answer.
Assuming that strings like '123456' and '123' are considered as not matching:
df_orig = pd.DataFrame(
{'col':['002446','002447-01','002447-02','002448','00244','002448']}
)
df = df_orig.reset_index(drop=True) # reset your index
first_6 = df['col'].str.slice(stop=6)
mask = first_6 != first_6.shift(fill_value=first_6[0])
df.index = df.index + mask.cumsum()
df = df.reindex(range(df.index[-1] + 1))
print(df)
col
0 002446
1 NaN
2 002447-01
3 002447-02
4 NaN
5 002448
6 NaN
7 00244
8 NaN
9 002448

Can I mix Python dataframe positional index and column name?

Can I get an entry in dataframe using negative/positional index and column name? For example,
import pandas as pd
df = pd.DataFrame({'a':[0,1,2], 'b':[3,4,5]})
# df is:
# a b
# 0 0 3
# 1 1 4
# 2 2 5
df.iloc[-1]['a'] # got 2
df['a'].iloc[-1] # got 2
df.loc[1,'a'] # got 1
df.loc[-1,'a'] # got KeyError: -1

No, use df.columns and get_loc, to convert column header label to index location:
df.iloc[-1, df.columns.get_loc('a')]
Output:
2

No you can't mix both within the same iloc/loc (this used to be the case with the now deprecated ix indexer)
The correct approach would be one of the double slice:
df.iloc[-1]['a']
Note that the "issue" is not specific to negative positional indices. When you do df.loc[1,'a'] you are not slicing per position, but per name. This would also fail with a KeyError if you had no 1 in the index.

Return value from row based on column value pandas

I have a data frame that has a column of lists of strings, I want to find the value of a colum in a row which is based on the value of another column
i.e
samples subject trial_num
0 ['aa','bb'] 1 1
1 ['bb','cc'] 1 2
I have ['bb','cc'] and I want to get the value from the trial_num column where this list equals the samples colum, in this case 2.

Given the search column (samples) contains a list, it makes thing a tiny bit more complicated.
In this case, the apply() function can be used to test the values, and return a boolean mask, which can be applied to the DataFrame to obtain the required value.
Example code:
df.loc[df['samples'].apply(lambda x: x == ['bb', 'cc']), 'trial_num']
Output:
1 2
Name: trial_num, dtype: int64
To only return the required value (2), simply append .iloc[0] to the end of the statement, as:
df.loc[df['samples'].apply(lambda x: x == ['bb', 'cc']), 'trial_num'].iloc[0]
>>> 2

Dropping row in pandas by index - have to add to index?

I have a pandas dataframe of 182 rows that comes from read_csv. The first column, sys_code, contains various alphanumeric codes. I want to drop ones that start with 'FB' (there are 14 of these). I loop through the dataframe, adding what I assume would be the index to a list, then try to drop by index using the list. But this doesn't work unless I add 18 to each index number.
Without adding 18, I get a list containing numbers from 84 - 97. When I try to drop the rows using this list for indexes, I get KeyError: '[84] not found in axis'. But when I add 18 to each number, it works fine, at least for this particular dataset. But why is this? Shouldn't i be the same as the index number?
fb = []
i = 0
df.reset_index(drop=True)
for x in df['sys_code']:
if x[:2] == 'FB':
fb.append(i+18) #works
fb.append(i) # doesn't work
i += 1
df.drop(fb, axis=0, inplace=True)

You could use Series.str.startswith. Here's an example:
df = pd.DataFrame({'col1':['some string', 'FBsomething', 'FB', 'etc']})
print(df)
col1
0 some string
1 FBsomething
2 FB
3 etc
You could remove those strings that do not start with FB using:
df[~df.col1.str.startswith('FB')]
col1
0 some string
3 etc

Creating series from a series of list using positional list memeber

I have a data frame with a string column. I need to create a new column with 3rd element after col1.split(' '). I tried
df['col1'].str.split(' ')[0]
but all I get is error.
Actually I need to turn col1 into multiple columns after spliting by " ".
What is the correct way to do this ?

Consider this df
df = pd.DataFrame({'col': ['Lets say 2000 is greater than 5']})
col
0 Lets say 2000 is greater than 5
You can split and use str accessor to get elements at different positions
df['third'] = df.col.str.split(' ').str[2]
df['fifth'] = df.col.str.split(' ').str[4]
df['last'] = df.col.str.split(' ').str[-1]
col third fifth last
0 Lets say 2000 is greater than 5 2000 greater 5

Another way is:
df["third"] = df['col1'].apply(lambda x: x.split()[2])