A certain dataframe, has a similar break with the one below:
import pandas as pd
df = pd.DataFrame({'name': ['John', 'Elvis', 'Gerrard', 'Pitty'],
'age': [22,23,24,25],
'document': [111,222,333,4444]})
How can I make a filter to return only the rows where the values in the document column are only 3 digits?
logdf.query('2 <= log10(document) < 3')
name age document
0 John 22 111
1 Elvis 23 222
2 Gerrard 24 333
df = pd.DataFrame({
'name': ['John', 'Elvis', 'Gerrard', 'Pitty'],
'age': [22, 23, 24, 25],
'document': [11, 222, 999, 1000]
})
df
name age document
0 John 22 11 # 2 digit number
1 Elvis 23 222 # 3 digit number
2 Gerrard 24 999 # 3 digit number | edge case
3 Pitty 25 1000 # 4 digit number | edge case
Let's get only 3 digit numbers
df.query('2 <= log10(document) < 3')
name age document
1 Elvis 23 222
2 Gerrard 24 999
Try // which will return the divisor of 1000, since 3 digit so the divisor should be 0
df[df.document//1000==0]
Out[474]:
name age document
0 John 22 111
1 Elvis 23 222
2 Gerrard 24 333
We convert to str then count the len
df[df.document.astype(str).str.len().eq(3)]
Out[476]:
name age document
0 John 22 111
1 Elvis 23 222
2 Gerrard 24 333
Update
df[df.document.astype(str).str.split('.').str[0].str.len().eq(3)]
// solution ;p Also,. what about floats in the astype(str) solution?
df[df.document.floordiv(100).between(1, 9)]
df[df.document.between(100, 999)]df[df.document.apply(lambda x: len(str(x)) == 3)]
This will work for positive and negative numbers in the document column:
df[df['document'].abs().astype(str).str.len() == 3]
