I have a dataframe like this:
id String
1 345 -456 -13 879
2 158 -926 -81 249 35 -4 -53 9
3 945 -506 -103
I want to sort it in desending order like this
id String
1 879 345 -13 -457
2 249 158 35 9 -4 -53 -81 -926
3 945 -103 -506
I tried this:
df['string'] = df['string'].str.split(' ').map(lambda x: ' '.join(sorted(x)))
The above function do some kind of sorting but not the way I want it.
First is necessary convert values to integers, sorting and then convert to strings back:
f = lambda x: ' '.join(map(str, sorted(map(int, x), reverse=True)))
#another solution
#f = lambda x: ' '.join(str(z) for z in sorted((int(y) for y in x), reverse=True))
df['string'] = df['string'].str.split().map(f)
print (df)
id string
0 1 879 345 -13 -456
1 2 249 158 35 9 -4 -53 -81 -926
2 3 945 -103 -506
Or:
f = lambda x: ' '.join(map(str, sorted(map(int, x.split()), reverse=True)))
df['string'] = df['string'].map(f)
ValueError: invalid literal for int() with base 10: '345'. The dtype of df['string'] is object
print (df.head().to_dict()) ?{'string': {0: '-4200.0 -135400.0 -148000.0 -58500.00000000001 -2800.0 -26899.999999999996 -51400.0 -33800.0 -18600.0 -65900.0 -23499.999999999996', 1: '-15200.0 -148000.0 -67700.0 -26000.0 -27300.0 -28599.999999999996 -25799.999999999996 -108199.99999999999 -19900.0 -31900.0 -89400.0 -63300.0', 2: '-31400.0 -12800.0 -1200.0', 3: '-32700.000000000004 -15999.999999999998 -18300.0 -200.0 -7000.0 -99.99999999999999', 4: '0.0 -8100.0 -4399.999999999999'}}