Sort dictionary according to keys
Question:
I have this dictionary:
but this dictionary is badly sorted, it should be:
x_11,
x_12,
x_13,
x_14,
x_15,
x_16,
x_17,
x_18,
x_19,
x_110,
x_21,
x_22,
x_23,
x_24,
x_25,
x_26,
x_27,
x_28,
x_29,
x_210,
x_31, x_32, x_33, x_34, x_35, x_36, x_37, x_38, x_39, x_310,
The first digit at x is the row number, the second digit at x is the column number.
Answers:
Idea is sorting by string before 3 letters and integers after 3 letters:
print (d)
{'x_11': 0, 'x_110': 1, 'x_12': 2, 'x_13': 3, 'x_21': 4, 'x_210': 5, 'x_22': 6, 'x_23': 7}
d = dict(sorted(d.items(), key=lambda x: (x[0][:3], int(x[0][3:]))))
print (d)
{'x_11': 0, 'x_12': 2, 'x_13': 3, 'x_110': 1, 'x_21': 4, 'x_22': 6, 'x_23': 7, 'x_210': 5}
Or if solution above not working and need split by _ it is possible sorting by first letter after separator and integer for another values:
d = dict(sorted(d.items(), key=lambda x: (x[0].split('_')[1][:1], int(x[0].split('_')[1][1:]))))
print (d)
{'x_11': 0, 'x_12': 2, 'x_13': 3, 'x_110': 1, 'x_21': 4, 'x_22': 6, 'x_23': 7, 'x_210': 5}
You can use sorted method two-times
result = dict(sorted(d.items(), key=lambda s: int(s[0].split("_")[1])))
result = dict(sorted(result.items(), key=lambda s: int(s[0].split("_")[1][0])))
Output:
{'x_11': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0,
'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_110': 0, 'x_21': 0, 'x_22': 0,
'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0,
'x_29': 0, 'x_210': 0, 'x_31': 0, 'x_32': 0, 'x_33': 0, 'x_310': 0}
Explanation
- The input dictionary:
d = {'x_11': 0, 'x_110': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0,
'x_15': 0, 'x_16': 0, 'x_17': 0, 'x_18': 0, 'x_19': 0,
'x_21': 0, 'x_210': 0, 'x_22': 0, 'x_23': 0, 'x_24': 0,
'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0, 'x_29': 0,
'x_31': 0, 'x_310': 0, 'x_32': 0, 'x_33': 0}
The aim is to reach x_11, x12, x_13, ..., x_110, x_21, ... x_310
-
My approach:
- First, I get integer part of each key (i.e. get 11 from
x_11) then sort
result = dict(sorted(d.items(), key=lambda s: int(s[0].split("_")[1])))
Output:
{'x_11': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0, 'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_21': 0, 'x_22': 0, 'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0, 'x_29': 0, 'x_31': 0, 'x_32': 0, 'x_33': 0, 'x_110': 0, 'x_210': 0, 'x_310': 0}
- Second, sort based on the first integer (i.e. get 1 from
x_11)
result = dict(sorted(result.items(), key=lambda s: int(s[0].split("_")[1][0])))
Output:
{'x_11': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0, 'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_110': 0, 'x_21': 0, 'x_22': 0, 'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0, 'x_29': 0, 'x_210': 0, 'x_31': 0, 'x_32': 0, 'x_33': 0, 'x_310': 0}
You can use natsort.
You must install it. I used pip.
pip install natsort
from natsort import natsorted as nat
nat(d.items())
I used the dictionary:
d = {'x_11':93.0,'x_110':32.0,'x_15':0.0}
The output is:
[('x_11', 93.0), ('x_15', 0.0), ('x_110', 32.0)]
To get it back to dictionary just add one more step:
newdict = dict(nat(d.items()))
print(newdict)
{'x_11': 93.0, 'x_15': 0.0, 'x_110': 32.0}
As you tagged pandas, let’s try creating a dataframe and sorting it by two columns; we can then cast it back as a dictionary.
#assuming your dictionary is caleld d.
#not re-recreating your value data from image so coded all to 0.
d1 = (
pd.DataFrame.from_dict(d, orient="index")
.reset_index(0)
.rename(columns={"index": "keys"})
)
d1 = d1.assign(
order1=d1["keys"].str.split("_", expand=True)[1].astype(int),
order2=d1["keys"].str.split("_", expand=True)[1].str[0].astype(int),
)
This yields :
keys 0 order1 order2
0 x_11 0 11 1
1 x_12 0 12 1
2 x_13 0 13 1
3 x_14 0 14 1
4 x_15 0 15 1
5 x_16 0 16 1
6 x_17 0 17 1
7 x_18 0 18 1
8 x_19 0 19 1
9 x_110 0 110 1
10 x_21 0 21 2
11 x_22 0 22 2
12 x_23 0 23 2
13 x_24 0 24 2
14 x_25 0 25 2
15 x_26 0 26 2
16 x_27 0 27 2
17 x_28 0 28 2
18 x_29 0 29 2
19 x_210 0 210 2
20 x_31 0 31 3
21 x_32 0 32 3
22 x_33 0 33 3
23 x_34 0 34 3
24 x_35 0 35 3
25 x_36 0 36 3
26 x_37 0 37 3
27 x_38 0 38 3
28 x_39 0 39 3
29 x_310 0 310 3
sorted_dict = d1.sort_values(['order2','order1']).set_index('keys')[0].to_dict()
{'x_11': 0,
'x_12': 0,
'x_13': 0,
'x_14': 0,
'x_15': 0,
'x_16': 0,
'x_17': 0,
'x_18': 0,
'x_19': 0,
'x_110': 0,
'x_21': 0,
'x_22': 0,
'x_23': 0,
'x_24': 0,
'x_25': 0,
'x_26': 0,
'x_27': 0,
'x_28': 0,
'x_29': 0,
'x_210': 0,
'x_31': 0,
'x_32': 0,
'x_33': 0,
'x_34': 0,
'x_35': 0,
'x_36': 0,
'x_37': 0,
'x_38': 0,
'x_39': 0,
'x_310': 0}
I have this dictionary:
but this dictionary is badly sorted, it should be:
x_11,
x_12,
x_13,
x_14,
x_15,
x_16,
x_17,
x_18,
x_19,
x_110,
x_21,
x_22,
x_23,
x_24,
x_25,
x_26,
x_27,
x_28,
x_29,
x_210,
x_31, x_32, x_33, x_34, x_35, x_36, x_37, x_38, x_39, x_310,
The first digit at x is the row number, the second digit at x is the column number.
Idea is sorting by string before 3 letters and integers after 3 letters:
print (d)
{'x_11': 0, 'x_110': 1, 'x_12': 2, 'x_13': 3, 'x_21': 4, 'x_210': 5, 'x_22': 6, 'x_23': 7}
d = dict(sorted(d.items(), key=lambda x: (x[0][:3], int(x[0][3:]))))
print (d)
{'x_11': 0, 'x_12': 2, 'x_13': 3, 'x_110': 1, 'x_21': 4, 'x_22': 6, 'x_23': 7, 'x_210': 5}
Or if solution above not working and need split by _ it is possible sorting by first letter after separator and integer for another values:
d = dict(sorted(d.items(), key=lambda x: (x[0].split('_')[1][:1], int(x[0].split('_')[1][1:]))))
print (d)
{'x_11': 0, 'x_12': 2, 'x_13': 3, 'x_110': 1, 'x_21': 4, 'x_22': 6, 'x_23': 7, 'x_210': 5}
You can use sorted method two-times
result = dict(sorted(d.items(), key=lambda s: int(s[0].split("_")[1])))
result = dict(sorted(result.items(), key=lambda s: int(s[0].split("_")[1][0])))
Output:
{'x_11': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0,
'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_110': 0, 'x_21': 0, 'x_22': 0,
'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0,
'x_29': 0, 'x_210': 0, 'x_31': 0, 'x_32': 0, 'x_33': 0, 'x_310': 0}
Explanation
- The input dictionary:
d = {'x_11': 0, 'x_110': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0, 'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_21': 0, 'x_210': 0, 'x_22': 0, 'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0, 'x_29': 0, 'x_31': 0, 'x_310': 0, 'x_32': 0, 'x_33': 0}
The aim is to reach x_11, x12, x_13, ..., x_110, x_21, ... x_310
-
My approach:
- First, I get integer part of each key (i.e. get 11 from
x_11) then sort
result = dict(sorted(d.items(), key=lambda s: int(s[0].split("_")[1])))Output:
{'x_11': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0, 'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_21': 0, 'x_22': 0, 'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0, 'x_29': 0, 'x_31': 0, 'x_32': 0, 'x_33': 0, 'x_110': 0, 'x_210': 0, 'x_310': 0}- Second, sort based on the first integer (i.e. get 1 from
x_11)
result = dict(sorted(result.items(), key=lambda s: int(s[0].split("_")[1][0])))Output:
{'x_11': 0, 'x_12': 0, 'x_13': 0, 'x_14': 0, 'x_15': 0, 'x_16': 0, 'x_17': 0, 'x_18': 0, 'x_19': 0, 'x_110': 0, 'x_21': 0, 'x_22': 0, 'x_23': 0, 'x_24': 0, 'x_25': 0, 'x_26': 0, 'x_27': 0, 'x_28': 0, 'x_29': 0, 'x_210': 0, 'x_31': 0, 'x_32': 0, 'x_33': 0, 'x_310': 0} - First, I get integer part of each key (i.e. get 11 from
You can use natsort.
You must install it. I used pip.
pip install natsort
from natsort import natsorted as nat
nat(d.items())
I used the dictionary:
d = {'x_11':93.0,'x_110':32.0,'x_15':0.0}
The output is:
[('x_11', 93.0), ('x_15', 0.0), ('x_110', 32.0)]
To get it back to dictionary just add one more step:
newdict = dict(nat(d.items()))
print(newdict)
{'x_11': 93.0, 'x_15': 0.0, 'x_110': 32.0}
As you tagged pandas, let’s try creating a dataframe and sorting it by two columns; we can then cast it back as a dictionary.
#assuming your dictionary is caleld d.
#not re-recreating your value data from image so coded all to 0.
d1 = (
pd.DataFrame.from_dict(d, orient="index")
.reset_index(0)
.rename(columns={"index": "keys"})
)
d1 = d1.assign(
order1=d1["keys"].str.split("_", expand=True)[1].astype(int),
order2=d1["keys"].str.split("_", expand=True)[1].str[0].astype(int),
)
This yields :
keys 0 order1 order2
0 x_11 0 11 1
1 x_12 0 12 1
2 x_13 0 13 1
3 x_14 0 14 1
4 x_15 0 15 1
5 x_16 0 16 1
6 x_17 0 17 1
7 x_18 0 18 1
8 x_19 0 19 1
9 x_110 0 110 1
10 x_21 0 21 2
11 x_22 0 22 2
12 x_23 0 23 2
13 x_24 0 24 2
14 x_25 0 25 2
15 x_26 0 26 2
16 x_27 0 27 2
17 x_28 0 28 2
18 x_29 0 29 2
19 x_210 0 210 2
20 x_31 0 31 3
21 x_32 0 32 3
22 x_33 0 33 3
23 x_34 0 34 3
24 x_35 0 35 3
25 x_36 0 36 3
26 x_37 0 37 3
27 x_38 0 38 3
28 x_39 0 39 3
29 x_310 0 310 3
sorted_dict = d1.sort_values(['order2','order1']).set_index('keys')[0].to_dict()
{'x_11': 0,
'x_12': 0,
'x_13': 0,
'x_14': 0,
'x_15': 0,
'x_16': 0,
'x_17': 0,
'x_18': 0,
'x_19': 0,
'x_110': 0,
'x_21': 0,
'x_22': 0,
'x_23': 0,
'x_24': 0,
'x_25': 0,
'x_26': 0,
'x_27': 0,
'x_28': 0,
'x_29': 0,
'x_210': 0,
'x_31': 0,
'x_32': 0,
'x_33': 0,
'x_34': 0,
'x_35': 0,
'x_36': 0,
'x_37': 0,
'x_38': 0,
'x_39': 0,
'x_310': 0}