Choice made by Python 3.5 to choose the keys when comparing them in a dictionary

Solution 1:

Dictionaries are implemented as hash tables and there are two important concepts when adding keys/values here: hashing and equality.

To insert a particular key/value, Python first computes the hash value of the key. This hash value is used to determine the row of the table where Python should first attempt to put the key/value.

If the row of the hash table is empty, great: the new key/value can inserted into the dictionary, filling the empty row.

However, if there's already something in that row, Python needs to test the keys for equality. If the keys are equal (using ==) then they're deemed to be the same key and Python just needs to update the corresponding value on that row.

(If the keys are not equal Python looks at other rows in the table until it finds the key or reaches an empty row, but that's not relevant for this question.)

When you write {True: 'yes', 1: 'No'}, you are telling Python to create a new dictionary and then fill it with two key/value pairs. These are processed left to right: True: 'yes' then 1: 'No'.

We have hash(True) equals 1. The key True goes in at row 1 in the hash table and the string 'yes' is its value.

For the next pair, Python sees that hash(1) is also 1 and so looks at row 1 of the table. There's something already there, so now Python checks the keys for equality. We have 1 == True so 1 is deemed to be the same key as True and so its corresponding value is changed to the string 'No'.

This results in a dictionary with one entry: {True: 'No'}.

If you want to peer at the guts of CPython 3.5 to see what creating a dictionary looks below the surface-Python level, here's more detail.

• The Python code {True: 'yes', 1: 'No'} is parsed into tokens and given to the compiler. Given the syntax, Python knows that a dictionary must be created using the values inside the braces. Byte code to load the four values onto the virtual machine's stack (LOAD_CONST) and then build the dictionary (BUILD_MAP) is queued up.

• The four constant values are pushed onto the top of the stack in the order that they're seen:
  
  

  'No'
  1
  'yes'
  True
  

• The opcode BUILD_MAP is then called with the argument 2 (Python counted two key/value pairs). This opcode is responsible for actually creating dictionary from the items on the stack. It looks like this:
  
  

  TARGET(BUILD_MAP) {
      int i;
      PyObject *map = _PyDict_NewPresized((Py_ssize_t)oparg);
      if (map == NULL)
          goto error;
      for (i = oparg; i > 0; i--) {
          int err;
          PyObject *key = PEEK(2*i);
          PyObject *value = PEEK(2*i - 1);
          err = PyDict_SetItem(map, key, value);
          if (err != 0) {
              Py_DECREF(map);
              goto error;
          }
      }
  
      while (oparg--) {
          Py_DECREF(POP());
          Py_DECREF(POP());
      }
      PUSH(map);
      DISPATCH();
  }
  

The three key steps here are as follows:

1. An empty hashtable is created using _PyDict_NewPresized. Small dictionaries (of just a few items, like 2 in this case) need a table with eight rows.

2. The for loop is entered, starting at 2 (in this case) and counting down to 0. PEEK(n) is a macro that points to the nth item down the stack. Therefore on the first iteration of the loop, we'll have

PyObject *key = PEEK(2*2);       /* item 4 down the stack */  
PyObject *value = PEEK(2*2 - 1); /* item 3 down the stack */

This means that *key will be True and *value will be 'yes' on the first loop through. On the second it will be 1 and 'No'.

3. PyDict_SetItem is called in each loop to put the current *key and *value into the dictionary. This is the same function that is called when you write dictionary[key] = value. It computes the hash of the key to work out where to look first in the hash table and then, if needed, compare the key to any existing key on that row (as discussed above).

Solution 2:

Basic premise is - True and 1 have same hashes and are equal to each other - that's why they cannot be separate keys in hash table (technically inequal object with same hashes may - but hash collisions decreases performance).

>>> True == 1
True
>>> hash(1)
1
>>> hash(True)
1

Now, let's consider a bytecode:

import dis
dis.dis("Dic = { True: 'yes', 1: 'No'}")

This prints:

      0 LOAD_CONST               0 (True)
      3 LOAD_CONST               1 ('yes')
      6 LOAD_CONST               2 (1)
      9 LOAD_CONST               3 ('No')
     12 BUILD_MAP                2
     15 STORE_NAME               0 (Dic)
     18 LOAD_CONST               4 (None)
     21 RETURN_VALUE

Basically what happens is that dict literal is tokenized to keys and values, and they are pushed to stack. After that BUILD_MAP 2 coverts two pairs of (keys, values) to dictionary.

Most likely order of data on stack (which seems to be determined by order of keys in dict literal) and implementation details of BUILD_MAP decides on resulting dict keys and values.

It seems like key-value assignments are done in order defined in dict literal. Assignment behaves same as d[key] = value operation, so it's basically:

• if key is not in dict (by equality): add key do dict
• store value under key

Given {True: 'yes', 1: 'No'}:

1. Start with {}

2. Add (True, 'yes')

   1. True is not in dict -> (True, hash(True)) == (True, 1) is new key in dict
   2. Update value for key equal to 1 to 'yes'

3. Add (1, 'no')

   1. 1 is in dict (1 == True) -> there is no need for new key in dictionary
   2. Update value for key equal to 1 (True) with value 'no'

Result: {True: 'No'}

As I commented, I do not know if this is guarranted by Python specs or if it is just CPython implementation-defined behavior, it may differ in other interpreter implementations.