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Create a nested dictionary for every distinct words in a list

I have a nested list, and for each list inside I want to create a dictionary that will contain another dictionary with the words related to a certain word as a key and the times they appear as the value. For example:
from
sentences = [["i", "am", "a", "sick", "man"],
["i", "am", "a", "spiteful", "man"],
["i", "am", "an", "unattractive", "man"],
["i", "believe", "my", "liver", "is", "diseased"],
["however", "i", "know", "nothing", "at", "all", "about", "my",
"disease", "and", "do", "not", "know", "for", "certain", "what", "ails", "me"]]
part of the dictionary returned would be:
{ "man": {"i": 3, "am": 3, "a": 2, "sick": 1, "spiteful": 1, "an": 1, "unattractive": 1}, "liver": {"i": 1, "believe": 1, "my": 1, "is": 1, "diseased": 1}...}
with as many keys as there are distinct words in the passage.
I've tried this:
d = {}
for row in sentences:
for words in rows:
if words not in d:
d[words] = 1
else:
d[words] += 1
But is only the way to count them, how could I use d as a value for another dictionary?

from collections import defaultdict
data = {}
for sentence in sentences:
for word in sentence:
data[word] = defaultdict(lambda: 0)
for sentence in sentences:
length = len(sentence)
for index1, word1 in enumerate(sentence):
for num in range(0, length - 1):
index2 = (index1 + 1 + num) % length
word2 = sentence[index2]
data[word1][word2] += 1
print(data)

sentences = [["i", "am", "a", "sick", "man"],
["i", "am", "a", "spiteful", "man"],
["i", "am", "an", "unattractive", "man"],
["i", "believe", "my", "liver", "is", "diseased"],
["however", "i", "know", "nothing", "at", "all", "about", "my",
"disease", "and", "do", "not", "know", "for", "certain", "what", "ails", "me"]]
# "as many keys as there are distinct words in the passage"
# Well then we need to start by finding the distinct words.
# sets always help for this.
# first we flatten the list. If you don't know what this is doing,
# search "flatten nested list Python". This is a common pattern:
flat_list = [term for group in sentences for term in group]
# now use set to find distinct words
distinct_words = set(flat_list)
# variable for final dictionary
result = {}
# define this function first. See invocation below
def find_related_counts(word):
# a nice way to do counts us with
# setdefault. If the term has already
# been counted, then it just increments.
# otherwise, it will create the key and
# initialise it to the default
related_counts = {}
for group in sentences:
# is "word" related to the terms in this group?
if word in group:
# yes it is! add the other terms:
for other in group:
# except, presumably, the word itself
if other != word:
related_counts.setdefault(other, 0)
related_counts[other] += 1
return related_counts
# for each word we have a key, and must find the value
for word in distinct_words:
# when dealing with nested anythings, it helps to
# make a function, so you don't have so much
# nesting in one place and separate things out
# nicely instead
value = find_related_counts(word)
result[word] = value
print(result)
print(result["man"])
OUTPUT:
{'spiteful': {'i': 1, 'am': 1, 'a': 1, 'man': 1}, 'and': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'unattractive': {'i': 1, 'am': 1, 'an': 1, 'man': 1}, 'nothing': {'however': 1, 'i': 1, 'know': 2, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'diseased': {'i': 1, 'believe': 1, 'my': 1, 'liver': 1, 'is': 1}, 'sick': {'i': 1, 'am': 1, 'a': 1, 'man': 1}, 'man': {'i': 3, 'am': 3, 'a': 2, 'sick': 1, 'spiteful': 1, 'an': 1, 'unattractive': 1}, 'do': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'believe': {'i': 1, 'my': 1, 'liver': 1, 'is': 1, 'diseased': 1}, 'i': {'am': 3, 'a': 2, 'sick': 1, 'man': 3, 'spiteful': 1, 'an': 1, 'unattractive': 1, 'believe': 1, 'my': 2, 'liver': 1, 'is': 1, 'diseased': 1, 'however': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'certain': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'what': 1, 'ails': 1, 'me': 1}, 'an': {'i': 1, 'am': 1, 'unattractive': 1, 'man': 1}, 'my': {'i': 2, 'believe': 1, 'liver': 1, 'is': 1, 'diseased': 1, 'however': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'a': {'i': 2, 'am': 2, 'sick': 1, 'man': 2, 'spiteful': 1}, 'am': {'i': 3, 'a': 2, 'sick': 1, 'man': 3, 'spiteful': 1, 'an': 1, 'unattractive': 1}, 'however': {'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'about': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'not': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'for': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'liver': {'i': 1, 'believe': 1, 'my': 1, 'is': 1, 'diseased': 1}, 'know': {'however': 1, 'i': 1, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'at': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'all': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'disease': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1, 'me': 1}, 'ails': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'me': 1}, 'me': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'what': 1, 'ails': 1}, 'what': {'however': 1, 'i': 1, 'know': 2, 'nothing': 1, 'at': 1, 'all': 1, 'about': 1, 'my': 1, 'disease': 1, 'and': 1, 'do': 1, 'not': 1, 'for': 1, 'certain': 1, 'ails': 1, 'me': 1}, 'is': {'i': 1, 'believe': 1, 'my': 1, 'liver': 1, 'diseased': 1}}
{'i': 3, 'am': 3, 'a': 2, 'sick': 1, 'spiteful': 1, 'an': 1, 'unattractive': 1}

Merge dicts in one and find mean

I am new in programming.
I have lists of dicts.
[{'Program Analysis': 0},
{'Algorithms': 0},
{'Number systems': 0},
{'Game theory': 1},
{'Algorithms': 1},
{'Number systems': 0},
{'Program Analysis': 0}]
I want to merge all dicts in one and then find mean.
{'Program Analysis': [0, 0]},
'Algorithms': [0, 1],
'Number systems': [0, 0],
'Game theory': 1 }
{'Program Analysis': 0},
'Algorithms': 0.5,
'Number systems': 0,
'Game theory': 1 }

def avg(num_list):
total = 0
for num in num_list:
total += num
return total * 1.0 / len(num_list)
my_list = [{'Program Analysis': 0}, {'Algorithms': 0}, {'Number systems': 0}, {'Game theory': 1}, {'Algorithms': 1}, {'Number systems': 0}, {'Program Analysis': 0}]
result = dict()
for d in my_list:
for name, num in d.items():
if name in result:
result[name].append(num)
else:
result[name] = [num]
for name, num_list in result.items():
result[name] = avg(num_list)

You can use a default dictionary to group the data and use mean function from statistics to calculate it:
from collections import defaultdict
from statistics import mean
data = [
{'Program Analysis': 0},
{'Algorithms': 0},
{'Number systems': 0},
{'Game theory': 1},
{'Algorithms': 1},
{'Number systems': 0},
{'Program Analysis': 0}
]
grouped_data = defaultdict(list)
for d in data:
for k, v in d.items():
grouped_data[k].append(v)
grouped_data = dict(grouped_data)
print(grouped_data)
>>> {'Program Analysis': [0, 0], 'Algorithms': [0, 1], 'Number systems': [0, 0], 'Game theory': [1]}
mean_data = {
k: mean(v) for k, v in grouped_data.items()
}
print(mean_data)
>>> {'Program Analysis': 0, 'Algorithms': 0.5, 'Number systems': 0, 'Game theory': 1}

Cleanest way to sum list of nested dicts

Is there a cleaner/more pythonic way of summing the contents of a list of nested dicts? Here's what I'm doing, but I suspect that there may be a better way:
list_of_nested_dicts = [{'class1': {'TP': 1, 'FP': 0, 'FN': 2}, 'class2': {'TP': 0, 'FP': 0, 'FN': 0}, 'class3': {'TP': 0, 'FP': 0, 'FN': 0}, 'class4': {'TP': 1, 'FP': 0, 'FN': 2}},
{'class1': {'TP': 1, 'FP': 0, 'FN': 2}, 'class2': {'TP': 0, 'FP': 0, 'FN': 0}, 'class3': {'TP': 0, 'FP': 0, 'FN': 0}, 'class4': {'TP': 1, 'FP': 0, 'FN': 2}},
{'class1': {'TP': 1, 'FP': 0, 'FN': 2}, 'class2': {'TP': 0, 'FP': 0, 'FN': 0}, 'class3': {'TP': 0, 'FP': 0, 'FN': 0}, 'class4': {'TP': 1, 'FP': 0, 'FN': 2}},
{'class1': {'TP': 1, 'FP': 0, 'FN': 2}, 'class2': {'TP': 0, 'FP': 0, 'FN': 0}, 'class3': {'TP': 0, 'FP': 0, 'FN': 0}, 'class4': {'TP': 1, 'FP': 0, 'FN': 2}}]
total_counts = {k:{'TP': 0, 'FP': 0, 'FN': 0} for k in list_of_nested_dicts[0].keys()}
for d in list_of_nested_dicts:
for label,counts_dict in d.items():
for k,v in counts_dict.items():
total_counts[label][k] += v
print(total_counts)
(Assuming all keys are exactly the same, but values could be any integer)

You can have a slightly tighter code using collections (similar result to #blhsing)
import collections
counts = collections.defaultdict(collections.Counter)
for d in list_of_nested_dicts:
for k, v in d.items():
counts[k].update(v)
This will give you a defaultdict of counters instead of only dicts, but they behave similarly. You can also explicitly cast them to dicts at the end if you want.
{'class1': {'FN': 8, 'FP': 0, 'TP': 4},
'class2': {'FN': 0, 'FP': 0, 'TP': 0},
'class3': {'FN': 0, 'FP': 0, 'TP': 0},
'class4': {'FN': 8, 'FP': 0, 'TP': 4}}
vs
defaultdict(<class 'collections.Counter'>,
{'class1': Counter({'FN': 8, 'TP': 4, 'FP': 0}),
'class2': Counter({'TP': 0, 'FP': 0, 'FN': 0}),
'class3': Counter({'TP': 0, 'FP': 0, 'FN': 0}),
'class4': Counter({'FN': 8, 'TP': 4, 'FP': 0})})

One thing in your code that stands out as "unclean" is the fact that you are hard-coding the keys of the sub-dicts in the initialization of total_counts. You can avoid such hard-coding by using the dict.setdefault and dict.get methods as you iterate over the items of the sub-dicts instead:
total_counts = {}
for d in list_of_nested_dicts:
for label, counts_dict in d.items():
for k, v in counts_dict.items():
total_counts[label][k] = total_counts.setdefault(label, {}).get(k, 0) + v

pandas - pd.replace and TypeError

I have all_data dataframe. I want to replace some categorical values in certain columns with numerical values. I'm trying to use this nested dictionary notation (I've checked that the brackets and curly brackets are in place, I don't think that's the issue):
all_data = all_data.replace({'Street': {'Pave': 1, 'Grvl': 0}},
{'LotShape': {'IR3': 1, 'IR2': 2, 'IR1': 3, 'Reg': 4}},
{'Utilities': {'ELO': 0, 'NoSeWa': 0, 'NoSewr': 0, 'AllPub': 1}},
{'LandSlope': {'Sev': 1, 'Mod': 2, 'Gtl': 3}},
{'ExterQual': {'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4, 'Ex': 5}},
{'ExterCond': {'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4, 'Ex': 5}},
{'BsmtQual': {'NA': 0, 'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4,'Ex': 5}},
{'BsmtCond': {'NA': 0, 'Po': 1, 'Fa': 2, 'TA': 3, 'Gd': 4,'Ex': 5}},
{'BsmtExposure': {'NA': 0, 'No': 1, 'Mn': 2, 'Av': 3, 'Gd': 4}},
{'BsmtFinType1': {'NA': 0, 'Unf': 1, 'LwQ': 2, 'Rec': 3, 'BLQ': 4, 'ALQ': 5, 'GLQ': 6}},
{'BsmtFinType2': {'NA': 0, 'Unf': 1,'LwQ': 2,'Rec': 3, 'BLQ': 4,'ALQ': 5, 'GLQ': 6}},
{'HeatingQC': {'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'CentralAir': {'No': 0,'Yes': 1}},
{'KitchenQual': {'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'Functional': {'Sal': -7,'Sev': -6,'Maj1': -5,'Maj2': -4,'Mod': -3,'Min2': -2,'Min1': -1,
'Typ': 0}},
{'FireplaceQu': {'NA': 0,'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'GarageFinish': {'NA': 0,'Unf': 1,'RFn': 2, 'Fin': 3}},
{'GarageQual': {'NA': 0, 'Po': 1,'Fa': 2, 'TA': 3,'Gd': 4, 'Ex': 5}},
{'GarageCond': {'NA': 0,'Po': 1,'Fa': 2,'TA': 3,'Gd': 4,'Ex': 5}},
{'PavedDrive': {'N': 0,'P': 0, 'Y': 1}},
{'Fence': {'NA': 0, 'MnWw': 1,'GdWo': 2,'MnPrv': 3,'GdPrv': 4}},
{'SaleCondition': {'Abnorml': 1, 'Alloca': 1, 'AdjLand': 1, 'Family': 1, 'Normal': 0,
'Partial': 0}}
)
Error:
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-40-f9c9c28b7237> in <module>()
22 {'Fence': {'NA': 0, 'MnWw': 1,'GdWo': 2,'MnPrv': 3,'GdPrv': 4}},
23 {'SaleCondition': {'Abnorml': 1, 'Alloca': 1, 'AdjLand': 1, 'Family': 1, 'Normal': 0,
---> 24 'Partial': 0}}
25 )
TypeError: replace() takes from 1 to 8 positional arguments but 23 were given
If I remove the 'SaleCondition' row from the above code, the error is again there but this time referring to 'Fence', and so on, for each line of code from bottom up. I've googled but have no idea what this means. Help MUCH appreciated.

You should do something like :
df.replace({'Fence':{'NA': 0, 'MnWw': 1,'GdWo': 2,'MnPrv': 3,'GdPrv': 4},'SaleCondition':{'Abnorml': 1, 'Alloca': 1, 'AdjLand': 1, 'Family': 1, 'Normal': 0,
'Partial': 0}})
the format should be .replace({'col1':{},'col2':{}}) not .replace({'col1':{}},{'col2':{}})

Django: Aggregation does not group by an integer

I want to count on an integer variable like I would with
select q001_r,count(q001_r) from ProjectData group by q001_r;
The result should be a dictionary saying I have e.g. 100 entries with a value of q001_r == 1
I have a really simple model
class ProjectData(models.Model):
"""No-SQL Flatfile der Variablen"""
caseid = models.PositiveIntegerField(unique=True)
...
q001_r = models.PositiveIntegerField()
and this is my aggregation funciton in the view
countInfosq1 = ProjectData.objects.values('q001_r').annotate(q1count=Count('q001_r'))
My result is not what I am expecting but:
[{'q1count': 1, 'q001_r': 1}, {'q1count': 1, 'q001_r': 3}, {'q1count': 1, 'q001_r': 1}, {'q1count': 1, 'q001_r': 3}, {'q1count': 1, 'q001_r': 2}, {'q1count': 1, 'q001_r': 2}, {'q1count': 1, 'q001_r': 2}, {'q1count': 1, 'q001_r': 1}, {'q1count': 1, 'q001_r': 3}]