map can be much faster than replace

If your dictionary has more than a couple of keys, using map can be much faster than replace. There are two versions of this approach, depending on whether your dictionary exhaustively maps all possible values (and also whether you want non-matches to keep their values or be converted to NaNs):

Exhaustive Mapping

In this case, the form is very simple:

df['col1'].map(di)       # note: if the dictionary does not exhaustively map all
                         # entries then non-matched entries are changed to NaNs

Although map most commonly takes a function as its argument, it can alternatively take a dictionary or series: Documentation for Pandas.series.map

Non-Exhaustive Mapping

If you have a non-exhaustive mapping and wish to retain the existing variables for non-matches, you can add fillna:

df['col1'].map(di).fillna(df['col1'])

as in @jpp's answer here: Replace values in a pandas series via dictionary efficiently

Benchmarks

Using the following data with pandas version 0.23.1:

di = {1: "A", 2: "B", 3: "C", 4: "D", 5: "E", 6: "F", 7: "G", 8: "H" }
df = pd.DataFrame({ 'col1': np.random.choice( range(1,9), 100000 ) })

and testing with %timeit, it appears that map is approximately 10x faster than replace.

Note that your speedup with map will vary with your data. The largest speedup appears to be with large dictionaries and exhaustive replaces. See @jpp answer (linked above) for more extensive benchmarks and discussion.

OP, the first dictionary will determine the naming and order of all columns, unless subsequent dictionaries introduce new keys.

All you have to do is pass your list of dictionaries to pandas and it will create the columns based on the dictionary keys automatically.

md = [
{'c':3,'a':1,'b':2},
{'d':44,'a':11,'b':22},
{'b':222}]

df=pd.DataFrame(md)

print(df.head())

     c     a    b     d
0  3.0   1.0    2   NaN
1  NaN  11.0   22  44.0
2  NaN   NaN  222   NaN

an alternative : changed name from dict to dicts to avoid confusion with the built-in type

df.set_index('key_col').T.fillna(dicts).T

           target_col
key_col 
   w         a
   c         B
   z         4

Another method with df.map and df.transform:

In [90]: mapping = {(0, 0) :'a', (0, 1) : 'b', (1, 0): 'c', (1, 1): 'd'}

In [91]: df['val'] = df.transform(lambda x: (x['q1'], x['q2']), axis=1).map(mapping); df
Out[91]: 
    q1  q2 val
0    0   1   b
1    0   1   b
2    0   1   b
3    0   1   b
4    0   1   b
5    0   1   b
6    0   1   b
7    0   1   b
8    0   1   b
9    0   1   b
10   1   1   d
11   1   1   d
12   0   1   b
13   0   1   b
14   1   0   c
15   0   0   a
16   0   0   a
17   0   0   a
18   0   0   a
19   0   0   a
20   0   0   a
21   0   0   a

You can also use zip to generate columns, apply pd.Series and then do the mapping:

In [119]: df['val'] = pd.Series(list(zip(df.q1, df.q2))).map(mapping); df
Out[119]: 
    q1  q2 val
0    0   1   b
1    0   1   b
2    0   1   b
3    0   1   b
4    0   1   b
5    0   1   b
6    0   1   b
7    0   1   b
8    0   1   b
9    0   1   b
10   1   1   d
11   1   1   d
12   0   1   b
13   0   1   b
14   1   0   c
15   0   0   a
16   0   0   a
17   0   0   a
18   0   0   a
19   0   0   a
20   0   0   a
21   0   0   a

Performance

jezrael's solution:

In [552]: %%timeit
     ...: idx = pd.MultiIndex.from_product([[0,1],[0,1]], names=('q1','q2'))
     ...: s = pd.Series(['a','b','c','d'], index=idx, name='val')
     ...: df.join(s, on=['q1','q2'])
     ...: 
100 loops, best of 3: 2.84 ms per loop

Proposed in this post:

In [553]: %%timeit
     ...: mapping = {(0, 0) :'a', (0, 1) : 'b', (1, 0): 'c', (1, 1): 'd'}
     ...: df.transform(lambda x: (x['q1'], x['q2']), axis=1).map(mapping)
     ...:  
1000 loops, best of 3: 1.7 ms per loop

list_to_change = [i for i in issue_short_form_map.keys()]

def check_replace(x):
    """
    Check elements in x and compare to change

    """
    for element_to_check in list_to_change:
        if x.__contains__(element_to_check):
            return x.replace(element_to_check,issue_short_form_map[element_to_check])
    return x
df["issue"]= df["issue"].map(check_replace)
print(df)
                        date  source destination     issue
0 2021-02-06 18:48:18.097962  London    New York    UP_LOW
1 2021-02-06 18:48:18.097962  Berlin       Tokyo   DN_HIGH
2 2021-02-06 18:47:08.209495   Paris     Toronto  DROP_LOW