Trying to find all occurrences of a substring within a string, and also keep n characters afterwards-CodePudding

For a dataframe, I am trying to extract all occurrences of "cash" and then n characters after them (which contains the cash amount). I have tried JSON, Regex, but they do not work as this dataframe is quite inconsistent.

So for example,

sample = pd.DataFrame({'LongString': ["I am trying to find out how much cash 15906810 
and this needs to be consistent cash :  69105060", 
"other words that are wrong cash : 11234 and more words cash 1526
"]})

And then my dataframe will look like

sample_resolved = pd.DataFrame({'LongString': ["I am trying to find out how much cash 15906810 
and this needs to be consistent cash :  69105060", 
"other words that are wrong cash : 11234 and more words cash 1526
"], 'cash_string' = ["cash  15906810 cash : 69105060", "cash : 11234 cash 1526]})

Each row of the dataframe is inconsistent. The ultimate goal is to create a new column that has all instances of "cash" followed by let's say 8-10 characters after it.

The ultimate goal would be to have a line that goes

df['cash_string'] = df['LongString'].str.findall('cash')

(but also includes the n characters after each 'cash' instance)

Thank you!

CodePudding user response：

In general, if there isn't a dataframe method (or combination thereof) that does what you're after, you can write a function that works on a single example and then pass it to the dataframe with series.apply(some_func).

So, a function that does what you're looking for:

def str_after_substr(s, substr='cash', offset=5):
    i = s.index(substr)
    start = i len(substr)
    return s[start:start offset]
# test
str_after_substr('moneymoneycashmoneyhoney')

# create the new column values and add it to the df
df['new_column] = df['old_column'].apply(str_after_substr)

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.apply.html

CodePudding user response：

Example

make minimal and reproducible example

df = pd.DataFrame(["abc cash : 1590 cde cash : 6910", "fgh cash : 1890 hij cash : 3410 cash : 4510"], columns=['col1'])

df

    col1
0   abc cash : 1590 cde cash : 6910
1   fgh cash : 1890 hij cash : 3410 cash : 4510

Code

s = df['col1'].str.extractall(r'(cash : \d )')[0]

s

  match
0  0        cash : 1590
   1        cash : 6910
1  0        cash : 1890
   1        cash : 3410
   2        cash : 4510
Name: 0, dtype: object

s.groupby(level=0).agg(', '.join)

0                 cash : 1590, cash : 6910
1    cash : 1890, cash : 3410, cash : 4510
Name: 0, dtype: object

Output

df.assign(col2=s.groupby(level=0).agg(', '.join))

    col1                                            col2
0   abc cash : 1590 cde cash : 6910                 cash : 1590, cash : 6910
1   fgh cash : 1890 hij cash : 3410 cash : 4510     cash : 1890, cash : 3410, cash : 4510

CodePudding user response：

To add on to @JCThomas 's answer, I'd change the str_after_substr function like below

def cash_finder(s, substr='cash', offset=10):
    ss = s.split(substr)
    cashlist = []
    for i in ss[1:]:
        cashlist.append(int(''.join([x for x in list(i[:offset].strip()) if re.match('\d',x) ])))
    return cashlist

This will give you all instances of cash in one sentence,

and, df operation will go like below.

ddf['cashstring'] = ddf['LongString'].apply(lambda x: [{'cash':i} for i in cash_finder(x)])