2️⃣ Dictionary Comprehension

Similarly we can get required dictionary from current dictionary using dictionary comprehension.
We have seen dictionary method .items() while understanding dictionary datatype,it returns the list of tuples containing key-value pairs.
Dictionary comprehension is optimal way and requires less number of lines as compared with using forloop logic.

Scenario 1: We have dictionary which contains keys as numbers & values as strings as follow. We want a new dictionary ‘len_dict’ which should have keys as numbers but values as length of the string.

💻 Example 1:

# implementation using for loop
string_dict = {1:'Hi', 2:'Bye', 3:'Morning', 4:'Hello'}

len_dict={}
for k in string_dict:
    len_dict[k] = len(string_dict[k])

print(len_dict)

✅ Output:

{1:2, 2:3, 3:7, 4:5}

Lets implement it using dictionary comprehension:

💻 Example 2:

# using dictionary comprehension
string_dict = {1:'Hi', 2:'Bye', 3:'Morning', 4:'Hello'}
len_dict = {k:len(v) for(k,v) in string_dict.items()}
print(len_dict)

✅ Output:

{1:2, 2:3, 3:7, 4:5}

Scenario 2: We have dictionary which represents division names as keys & no. of students as values. We want new dictionary which only contains the division keys which are having more than 100 students as values.

💻 Example 1:

# implementation using for loop

division_count = {'divA':35, 'divB':79, 'divC':137, 'divD':150, 'divE':99, 'divF':175}
required_dict = {}
for k in division_count:
    if division_count[k]>100:
        required_dict[k]=division_count[k]
print(required_dict)

✅ Output:

{'divC':137, 'divD':150, 'divF':175}

Lets increment it using dictionary comprehension:

💻 Example 2:

# implementation using dictionary comprehension

division_count = {'divA':35, 'divB':79, 'divC':137, 'divD':150, 'divE':99, 'divF':175}
required_dict = {k:v for (k,v) in division_count.items() if v>100}
print(required_dict)

✅ Output:

{'divC':137, 'divD':150, 'divF':175}

Be prepared with below questions prior to your interview !