Python: Dictionaries (Solutions)ΒΆ
Solutions:
Solution:
empty_dict = {} print empty_dict print len(empty_dict) # 0
Solution:
pronouns = {} pronouns[1] = "I" pronouns[2] = "you" pronouns[3] = "he" pronouns[4] = "we" pronouns[5] = "you" pronouns[6] = "they"
or:
pronouns = { 1: "I", 2: "you", 3: "he", 4: "we", 5: "you", 6: "them", }
Solution:
tenfold_of = {1: 10, 2: 20, 3: 30, 4: 40, 5: 50} print [tenfold_of[n] for n in range(2, 5)] print [tenfold_of[key] for key in tenfold_of.keys()]
Solution:
info_2TMV = { "pdb_id": "2TMV", "uniprot_id": "P69687 (CAPSD_TMV)", "number_scp_domains": 1, "number_pfam_domains": 1, }
Solution:
relatives_of = { "JULIE": ["FRANCIS", "BENEDICT"], "FRANCIS": ["JULIE", "MATTHEW"], "MATTHEW": ["FRANCIS", "BENEDICT"], "BENEDICT": ["JULIE", "MATTHEW"], } num_relatives_of_julie = len(relatives_of["JULIE"]) print num_relatives_of_julie
Solution:
from_2_bit_to_integer = { (0, 0): 0, (0, 1): 1, (1, 0): 2, (1, 1): 3, }
Note that we cannot use lists as keys: lists are not immutable!
Let’s print the value corresponding to 1, 0:
print from_2_bit_to_integer[(1, 0)] # ^^^^^^ # tuple
Solution:
ratios = { ("A", "T"): 10.0 / 3.0, ("A", "C"): 10.0 / 7.0, ("A", "G"): 10.0 / 6.0, ("T", "C"): 3.0 / 7.0, ("T", "G"): 3.0 / 6.0, ("C", "G"): 7.0 / 6.0, } print len(ratios) # 6 print len(ratios.keys()) # 6 print len(ratios.values()) # 6 print len(ratios.items()) # 6 # they all count the number of key-value pairs! # let's pront the keys of the dictionary, to get an idea print ratios.keys() # a list of tuples! print type(ratios.keys()) # list print type(ratios.keys()[0]) # tuple contains_T_A = ("T", "A") in ratios.keys() print contains_T_A # False contains_C_G = ("C", "G") in ratios.keys() print contains_C_G # True # let's do the same with has_key() print ratios.has_key(("T", "A")) # False print ratios.has_key(("C", "G")) # True # let's print the values of the dictionary to get an idea print ratios.values() # a list of int! print type(ratios.values()[0]) # int contains_2 = 2 in ratios.values() print contains_2 # True contains_3 = 3 in ratios.values() print contains_3 # False # let's print the key-value pairs to get an idea print ratios.items() # it's a list of pairs (tuple): the first element, the key, is # a tuple itself, the second is an int print (("A", "T"), 2) in ratios.items() # True print (("C", "G"), 1000) in ratios.items() # False # list comprehension: keys = [key_value[0] for key_value in ratios.items()] values = [key_value[-1] for key_value in ratios.items()]
Solution:
map = { "zero": 1, "one": 2, "two": 4, "three": 8, "four": 16, "five": 32, } # all the keys of map are strings so keys() returns a list of strings # we can use directly join() string_of_keys = " ".join(map.keys()) # the values of map are int, so we cannot use directly join() # we first have to transform all values from int to string string_of_values = " ".join(str(value) for value in map.values()) list_of_keys = map.keys() print list_of_keys # not ordered ordered_list_of_keys = map.keys() ordered_list_of_keys.sort() print ordered_list_of_keys # now it is ordered list_of_values_ordered_by_keys = \ [map[key] for key in ordered_list_of_keys]
Solution:
# using a list comprehension we can apply the dictionary to # the list: it's necessary to use lower() *before* # using an aminoacid as a key of translation_of! translation= [translation_of[aa.lower()] for aa in list] print translation # now we can use join() to concatenate the various parts result = " ".join(translation) print result
or, in a single line:
print " ".join([translation_of[aa.lower()] for aa in list])
Solution:
propensities = { 'N': 0.2299, 'P': 0.5523, 'Q': -0.1877, 'A': -0.2615, 'R': -0.1766, 'S': 0.1429, 'C': -0.01515, 'T': 0.0089, 'D': 0.2276, 'E': -0.2047, 'V': -0.3862, 'F': -0.2256, 'W': -0.2434, 'G': 0.4332, 'H': -0.0012, 'Y': -0.2075, 'I': -0.4222, 'K': -0.100092, 'L': 0.33793, 'M': -0.22590 } # using a list comprehension pos_aa = [aa for aa in propensities.keys() if propensities[aa]>0] print pos_aa