You can use applymap:

df = pd.DataFrame({'nearby_subway_station':['yes','no'], 'Station':['no','yes']})
print (df)
  Station nearby_subway_station
0      no                   yes
1     yes                    no

dict_map_yn_bool={'yes':True, 'no':False}

df = df.applymap(dict_map_yn_bool.get)
print (df)
  Station nearby_subway_station
0   False                  True
1    True                 False

Another solution:

for x in df:
    df[x] = df[x].map(dict_map_yn_bool)
print (df)
  Station nearby_subway_station
0   False                  True
1    True                 False

Thanks Jon Clements for very nice idea - using replace:

df = df.replace({'yes': True, 'no': False})
print (df)
  Station nearby_subway_station
0   False                  True
1    True                 False

Some differences if data are no in dict:

df = pd.DataFrame({'nearby_subway_station':['yes','no','a'], 'Station':['no','yes','no']})
print (df)
  Station nearby_subway_station
0      no                   yes
1     yes                    no
2      no                     a

applymap create None for boolean, strings, for numeric NaN.

df = df.applymap(dict_map_yn_bool.get)
print (df)
  Station nearby_subway_station
0   False                  True
1    True                 False
2   False                  None

map create NaN:

for x in df:
    df[x] = df[x].map(dict_map_yn_bool)

print (df)
  Station nearby_subway_station
0   False                  True
1    True                 False
2   False                   NaN

replace dont create NaN or None, but original data are untouched:

df = df.replace(dict_map_yn_bool)
print (df)
  Station nearby_subway_station
0   False                  True
1    True                 False
2   False                     a

Keep this paradigm in your head: dataframes are meant to be instantiated from a data source, they aren't meant to be created as a blank sheet and 'filled in' later on. So also in this case: don't build a dataframe upfront, instead create the dataframe using the dict as the basis for its data. From Create a Pandas DataFrame from List of Dicts:

cols = ['timestamp', 'name', 'adress', 'phone', 'website', 'rating', 'number_of_ratings', 'type']
df = pd.DataFrame(multidict, columns=cols)

Two steps:

First, let's turn those None values to NaNs so we can use numeric operations.

df['category'] = pd.to_numeric(df['category']) # add arg ,errors='coerce' if needed.

Second, let's use groupby transform and max to fill only the NaN values:

df["category"] = df["category"].fillna(
    df.groupby(["company", "product"])["category"].transform("max")
)
print(df)

    company    product  category
0  Company1  Product A       1.0
1  Company1  Product A       2.0
2  Company1  Product F       3.0
3  Company1  Product A       2.0
4  Company2  Product F       5.0
5  Company2  Product F       5.0

df['Gender'] = df.Name.map(lambda x: d[x][0])
df['Age'] = df.Name.map(lambda x: d[x][1])

You can create a MultiIndex from two series and then map. Data from @ALollz.

df['CountyType'] = df.set_index(['County', 'State']).index.map(dct.get)

print(df)

  County  State CountyType
0      A      1        One
1      A      2       None
2      B      1       None
3      B      2        Two
4      B      3      Three

For improve performance in larger DataFrames is possible use Series with DataFrame.join:

df = pd.DataFrame(df)

df1 = df.join(pd.Series(d, name='new'), on=[0,1,2])
print (df1.head(10))
     0   1     2  new
0  EFG  DS   321    1
1  EFG  DS   900   51
2  EFG  DS   900   51
3  EFG   Q   900   51
4  EFG  DS  1000   41
5  EFG  DS  1000   41
6  EFG  DS  1000   41
7  ABC  DS   444  421
8  EFG  DS   900   51
9  EFG  DS   900   51

Another idea (similar solution like in question):

 res = [d[(a,b,c)] for a,b,c in zip(df[0], df[1], df[2])]

Or:

 res = [d[(a,b,c)] for a,b,c in df[[0,1,2]].to_numpy()]

Using .apply

Ex:

df['Rate'] = df.apply(lambda x: rates[x['Currency']][x['Year']], axis=1)
# OR
df['Rate'] = df.apply(lambda x: rates.get(x['Currency'], dict()).get(x['Year'], 1), axis=1)
print(df)

Output:

  Item Currency  Year  Rate
0    1      USD  2019     1
1    2      USD  2020     2
2    3      CAD  2021     6
3    4      CAD  2019     4
4    5      GBP  2020     1

Convert your dictionary to Series and perform a simple comparison:

df.ge(pd.Series(checkdict)).replace({True: 'yes', False: 'no'})

output:

  col1 col2 col3
0   no   no  yes
1   no   no  yes
2  yes  yes  yes

To get aggregation per row:

df['any'] = df.ge(pd.Series(checkdict)).any(1).map({True: 'yes', False: 'no'})

output:

   col1  col2  col3  any
0     1     2   3.2  yes
1     2     3   4.2  yes
2     3     4   7.7  yes

Try with map then str.split

df[['val1', 'val2', 'val3']] = df['key'].map(d1).str.split(', ',expand=True).values
df
    key   val1   val2   val3
0  1231   val1   val2   val3
1  1232  val1a  val2a  val3a

You can use .replace. For example:

>>> df = pd.DataFrame({'col2': {0: 'a', 1: 2, 2: np.nan}, 'col1': {0: 'w', 1: 1, 2: 2}})
>>> di = {1: "A", 2: "B"}
>>> df
  col1 col2
0    w    a
1    1    2
2    2  NaN
>>> df.replace({"col1": di})
  col1 col2
0    w    a
1    A    2
2    B  NaN

or directly on the Series, i.e. df["col1"].replace(di, inplace=True).