Debayan Chakraborty - 11 months ago 59

Python Question

i have an input to coloumns of a dataframe as 12345 and want to output to excel sheet as 1XXX5 how to do this . The data type in the dataframe coloumn is an integer

`df=pd.read_excel('INVAMF.xls',sheetname=4,header=0,skiprows=0)`

#df created

print df.dtypes

print np.count_nonzero(pd.value_counts(df['ACCOUNT_NUMBER'].values))

s = (df['ACCOUNT_NUMBER'])

print s

s = s.astype(str)

s.apply(lambda x: x[0] + 'X' * (len(x) - 2) + x[-1])

print s

0 32642

1 32643

2 32644

3 32677

4 32680

5 32680

6 32688

7 32688

8 32695

9 32708

10 32708

11 32709

12 32710

13 32734

14 32734

15 32738

16 32738

17 6109

18 6109

`List item`

Answer Source

As you've failed to post any data and code here is a general form assuming that numbers are varying length:

```
In [141]:
s = pd.Series([8815392,2983])
s = s.astype(str)
s.apply(lambda x: x[0] + 'X' * (len(x) - 2) + x[-1])
Out[141]:
0 8XXXXX2
1 2XX3
dtype: object
```

if the numbers are equal length you can use a vectorised method to set the entire column:

```
In [142]:
s = pd.Series([8815392,1291283])
s = s.astype(str)
s.str[0] + 'X' * (s.str.len() - 2)[0] + s.str[-1]
Out[142]:
0 8XXXXX2
1 1XXXXX3
dtype: object
```

Also just to clarify a common problem you need to assign back the operation as most pandas methods return a copy and don't work in place, although some methods do have a `inplace`

arg. So you need to do the following:

```
s = s.apply(lambda x: x[0] + 'X' * (len(x) - 2) + x[-1])
```