x = ['1', '2', '4', 'c'], so x[1]=='2', which makes the expression (x[0] != "1" and x[1] != "2" and x[2] != "3") be evaluated as False.

When conditions are joined by and, they return True only if all conditions are True, and if they are joined by or, they return True when the first among them is evaluated to be True.

You need () instead []:

arrival_delayed_weather = (flight_data_finalcopy["ArrDelay"] > 0) & 
                           (flight_data_finalcopy["WeatherDelay"]>0)

But it seems you need ix for selecting columns UniqueCarrier and AirlineID by mask - a bit modified boolean indexing:

mask = (flight_data_finalcopy["ArrDelay"] > 0) & 
        (flight_data_finalcopy["WeatherDelay"]>0)
arrival_delayed_weather_filter=flight_data_finalcopy.ix[mask, ["UniqueCarrier","AirlineID"]]

Sample:

flight_data_finalcopy = pd.DataFrame({'ArrDelay':[0,2,3],
                                      'WeatherDelay':[0,0,6],
                                      'UniqueCarrier':['s','a','w'],
                                      'AirlineID':[1515,3546,5456]})

print (flight_data_finalcopy)
   AirlineID  ArrDelay UniqueCarrier  WeatherDelay
0       1515         0             s             0
1       3546         2             a             0
2       5456         3             w             6

mask = (flight_data_finalcopy["ArrDelay"] > 0) & (flight_data_finalcopy["WeatherDelay"]>0)
print (mask)
0    False
1    False
2     True
dtype: bool

arrival_delayed_weather_filter=flight_data_finalcopy.ix[mask, ["UniqueCarrier","AirlineID"]]
print (arrival_delayed_weather_filter)
  UniqueCarrier  AirlineID
2             w       5456

Use () because operator precedence:

temp2 = df[~df["Def"] & (df["days since"] > 7) & (df["bin"] == 3)]

Alternatively, create conditions on separate rows:

cond1 = df["bin"] == 3    
cond2 = df["days since"] > 7
cond3 = ~df["Def"]

temp2 = df[cond1 & cond2 & cond3]

Sample:

df = pd.DataFrame({'Def':[True] *2 + [False]*4,
                   'days since':[7,8,9,14,2,13],
                   'bin':[1,3,5,3,3,3]})

print (df)
     Def  bin  days since
0   True    1           7
1   True    3           8
2  False    5           9
3  False    3          14
4  False    3           2
5  False    3          13


temp2 = df[~df["Def"] & (df["days since"] > 7) & (df["bin"] == 3)]
print (temp2)
     Def  days since  bin
3  False          14    3
5  False          13    3

Another idea is use functions like Series.lt Series.gt Series.le Series.ge Series.ne:

temp2 = df[~df["Def"] & df["days since"].gt(7) & df["bin"].eq(3)]
print (temp2)
     Def  days since  bin
3  False          14    3
5  False          13    3

The logic behind the code is wrong.

If you do a_list['type2'] != unwanted_type2 is like a_list['type2'] !={'Apple','Banana','Orange','Melon'}, and you are comparing a value to the entire unwated_type2.

To resolve it, you can use not in in the condition:

B = []

wanted_type1 = 'A'
unwanted_type2 = {'Apple','Banana','Orange','Melon'}
unwanted_type3 = {'stage','books','films','music'}

a_list = some_variable['response']['results']
for list in a_list:
    if (a_list['type1'] == 'A') and (a_list['type2'] not in unwanted_type2) and (a_list['typeb'] not in unwanted_type3):
       B.append(list['type4'])

With this method you are looking at

Just transform the and inside the lamda, and combine your conditions there:

def check_odd(x):
    return (x % 2) == 0

l1 = list(range(1,20))
l2 = list(filter((lambda x:x>=5 and check_odd(x)), l1))
print(l2)

Update
You can actually combine any number of conditions from other functions by wrapping them all inside the lambda. Here's an example:

def check_odd(x):
    return (x % 2) == 0

def check_greater_than_five(x):
    return x >= 5

def check_divisible_by_three(x):
    return x % 3 == 0

l1 = list(range(1,20))
l2 = list(filter((lambda x: (check_odd(x) or check_divisible_by_three(x)) and check_greater_than_five(x)), l1))
print(l2)   #prints all numbers that are greater than five, and are divisible by either 2 or 3

You need:

fil_1 = test['col_a'].isin(['abc','def','ghi'])
fil_2 = test['col_b'].isin(['yes'])
fil_3 = test['col_c'].isin(['a'])

or

test.isin({'col_a': ['abc','def','ghi'],
           'col_b': ['yes'],
           'col_c' :['a']}).all(axis = 1)

df_filtered = test[fil_1 & fil_2 & fil_3]
print(df_filtered)
   col_a col_b col_c
0    abc   yes     a
2    abc   yes     a
4    def   yes     a
6    def   yes     a
8    ghi   yes     a
10   ghi   yes     a

or logic |

fil = test.isin({'col_a': ['abc','def','ghi'],'col_b': ['yes'],'col_c' :['a']})
df_filtered = df[fil]
print(df_filtered)

   col_a col_b col_c
0    abc   yes     a
1    abc   NaN   NaN
2    abc   yes     a
3    def   NaN   NaN
4    def   yes     a
5    def   NaN   NaN
6    def   yes     a
7    def   NaN   NaN
8    ghi   yes     a
9    ghi   NaN   NaN
10   ghi   yes     a

Now if we also use DataFrame.all:

df_filtered = df[fil.all(axis = 1)]
print(df_filtered)
   col_a col_b col_c
0    abc   yes     a
2    abc   yes     a
4    def   yes     a
6    def   yes     a
8    ghi   yes     a
10   ghi   yes     a

Detail

print(fil)
    col_a  col_b  col_c
0    True   True   True
1    True  False  False
2    True   True   True
3    True  False  False
4    True   True   True
5    True  False  False
6    True   True   True
7    True  False  False
8    True   True   True
9    True  False  False
10   True   True   True

print(test.isin({'col_a': ['abc','def','ghi']}))
    col_a  col_b  col_c
0    True  False  False
1    True  False  False
2    True  False  False
3    True  False  False
4    True  False  False
5    True  False  False
6    True  False  False
7    True  False  False
8    True  False  False
9    True  False  False
10   True  False  False

this return False in columns differences than col_a so you got NaN values ​​because you were using &

You can filter your list:

list_of_stuff = [
    "aus-airport-1",
    "aus-airport-2",
    "us-airport-1",
    "us-airport-2",
    "aus-ship-1",
    "us-ship-99",
    "nz-airport-1"
]

is_airport = lambda x: "-airport-" in x
is_ship = lambda x: "-ship-" in x

airports_excluding_nz = lambda x: is_airport(x) and not x.startswith("nz-")
airports_in_aus = lambda x: is_airport(x) and x.startswith("nz-")
ships = lambda x: is_ship(x)

print ("all regions excluding nz:" , 
       ", ".join( filter(airports_excluding_nz , list_of_stuff) ) )
print ("all regions in aus:", 
       ", ".join( filter(airports_in_aus, list_of_stuff) ) )
print ("all ships:", 
       ", ".join( filter(ships, list_of_stuff) ) )

Check results:

all regions excluding nz aus-airport-1, aus-airport-2, us-airport-1, us-airport-2
all regions in aus nz-airport-1
all ships aus-ship-1, us-ship-99

This should work. (after a few edits)

bigrams = ["st", "te", "ev", "on"]
df['count'] = df.Names.apply(lambda x: sum([1 for v in bigrams if v in x]))
df[df['count'] >= 3].Names