In this simple case, you can create a dictionary with the keys being the roles themselves: > roles = {d[‘role’]:d for d in dic} > roles[‘NOC’][’email’] Note that you are still looping through all dictionary items to create the new one. 0 solved Python dictionary printing specific value [closed]
@StevenRumbalski has given the right answer i.e., output = filter(lambda k: 5 <= dict1[k] <= 8, dict1) solved Use filter method and find the keys with values between 5 to 8 from the given dictionary
def intersect(a, b): a, b = set(a), set(b) return list(a & b) def merge_dicts(d1, d2): return {k: intersect(v1, v2) for (k, v1), v2 in zip(d1.items(), d2.values())} dictionary_1 = {‘A’ :[‘B’,’C’,’D’],’B’ :[‘A’]} dictionary_2 = {‘A’ :[‘A’,’B’], ‘B’ :[‘A’,’F’]} merge = merge_dicts(dictionary_1, dictionary_2) print(merge) # {‘A’: [‘B’], ‘B’: [‘A’]} 4 solved How to merge two dictionaries with … Read more
sample = “i am feeling good” output = {} output[“word”] = len(sample.split()) output[“chars”] = len(list(sample)) for elem in sample.split(): output[elem] = len(list(elem)) print(output) output: {‘word’: 4, ‘chars’: 17, ‘i’: 1, ‘am’: 2, ‘feeling’: 7, ‘good’: 4} solved A dictionary that returns number of word and each word’s lenght Python
No need to bother with pandas. Text files are iterable. Just open it, operate on the line (string) and fill a dictionnary. file = “font.txt” with open(file, “r”) as f: dic = dict() for line in f: x = line.strip(“\n”).split(” “) key = int(x[0].strip(“px”)) value = int(x[1]) if key not in dic.keys(): dic[key] = [value] … Read more
Assuming I understood your question, these should do it: for person in sorted(cat1): print(person, max(cat1.get(person))) result: ben 9 jeff 6 sam 9 then: for person in sorted(cat1, key=lambda x: max(cat1.get(x)), reverse=True): print(person, max(cat1.get(person))) result: ben 9 sam 9 jeff 6 then: for person in sorted(cat1, key=lambda x: sum(cat1.get(x))/len(cat1.get(x)), reverse=True): print(person, sum(cat1.get(person))/len(cat1.get(person))) result: ben 7.666666666666667 sam … Read more
type casting in golang is done by appending .(T) ip := cookieData[“ip”].(map[string]string) email := ip[“email”] 2 solved Get a single key value from map[string]string
Here is an informal explaination: The map() method, when applied to a collection of one type of thing (e.g. a collections of lines in a file) and provided a function (e.g. extract the second and third item from the given string) will return a collection of the results of applying that function to each item … Read more
Your dict is the wrong way round. The thing you want to look up (key) goes first in each pair pCode = {“T”: “Alberta”} a = input(“Enter a 6 character postal code: “) You can get the first character of the post code as a[0] print(pCode.get(a[0])) solved First character in a dictionary Python [closed]
Just loop through your dictionary elements, reformatting the list of lists into list of dictionary using a list comprehension. original = { ‘t0’: [[‘cat0’, [‘eagle0’]], [‘cat1’, [‘eagle1’]]], ‘t1’: [[‘cat2’, [‘eagle2’, ‘eagle3’]]] } result = [] for key, cats in original.items(): cats = [{‘cat’: cat, ‘eagles’: eagles} for cat, eagles in cats] result.append({‘t’: key, ‘cats’: cats}) … Read more
Try using ast.literal_eval: import ast s=”{(76034,0),(91316,0),(221981,768),(459889,0),(646088,0)}” print(dict(ast.literal_eval(‘[‘ + s.strip(‘{}’) + ‘]’))) Output: {76034: 0, 91316: 0, 221981: 768, 459889: 0, 646088: 0} solved Convert a string representation of a set of tuples to a dictionary [duplicate]
@Override public void onGetSuggestions(final SuggestionsInfo[] arg0) { isSpellCorrect = false; final StringBuilder sb = new StringBuilder(); for (int i = 0; i < arg0.length; ++i) { // Returned suggestions are contained in SuggestionsInfo final int len = arg0[i].getSuggestionsCount(); if(editText1.getText().toString().equalsIgnoreCase(arg0[i].getSuggestionAt(j)) { isSpellCorrect = true; break; } } } You can find the whole project from this … Read more
dictA = {“f1” : [“m”], “f2” : [“p”,”y”,”o”,”a”,”s”,”d”,”f”,”g”], “f3” : [“w”,”t”], “f5” : [“z”,”x”], “f6” : [“c”,”v”]} result = [] limit_size = 3 values_list = [] for values in dictA.itervalues(): values_list.append(len(values)) for i in range(0,max(values_list)): keys = list(dictA.keys()) count = 0 while count < len(keys): try: result.append(dictA[keys[count]][i]) except IndexError: dictA.pop(keys[count]) count = count + 1 … Read more
Are you trying to write your dictionary in a Excel Spreadsheet? In this case, you could use win32com library: import win32com.client xlApp = win32com.client.DispatchEx(‘Excel.Application’) xlApp.Visible = 0 xlBook = xlApp.Workbooks.Open(my_filename) sht = xlBook.Worksheets(my_sheet) row = 1 for element in dict.keys(): sht.Cells(row, 1).Value = element sht.Cells(row, 2).Value = dict[element] row += 1 xlBook.Save() xlBook.Close() Note that … Read more
Hints I would follow this steps: Create a list where I will store my partial strings Start iterating the string Store the initial position and the current character Keep iterating until the character is different Store in the list the partial string from the initial position you stored until 1 less than the current position … Read more