list.append copies the last item only - python

This might endup in very silly question, but being a newbie in python i am not able to find a good solution to following problem.
class Preprocessor:
mPath = None;
df = None;
def __init__(self, path):
self.mPath = path;
def read(self):
self.df = pd.read_csv(self.mPath);
return self.df;
def __findUniqueGenres(self):
setOfGenres = set();
for index, genre in self.df['genres'].iteritems():
listOfGenreInMovie = genre.lower().split("|");
for i, _genre in np.ndenumerate(listOfGenreInMovie):
setOfGenres.add(_genre)
return setOfGenres;
def __prepareDataframe(self, genres):
all_columns = set(["title", "movieId"]).union(genres)
_df = pd.DataFrame(columns=all_columns)
return _df;
def __getRowTemplate(self, listOfColumns):
_rowTemplate = {}
for col in listOfColumns:
_rowTemplate[col] = 0
return _rowTemplate;
def __createRow(self, rowTemplate, row):
rowTemplate['title'] = row.title;
rowTemplate['movieId'] = row.movieId;
movieGenres = row.genres.lower().split("|");
for movieGenre in movieGenres:
rowTemplate[movieGenre] = 1;
return rowTemplate;
def tranformDataFrame(self):
genres = self.__findUniqueGenres();
print('### List of genres...', genres);
__df = self.__prepareDataframe(genres); # Data frame with all required columns.
rowTemplate = self.__getRowTemplate(__df.columns)
print('### Row template looks like -->', rowTemplate)
collection = []
for index, row in self.df.iterrows():
_rowToAdd=self.__createRow(rowTemplate, row);
print('### Row looks like', _rowToAdd)
collection.append(_rowToAdd)
print('### Collection looks like', collection)
return __df.append(collection)
Here when i am trying to append a _rowToAdd to collection, it endsup having a collection of last item ( last row of self.df).
Below are logs for the same (self.df has 3 rows here),
### List of genres... {'mystery', 'horror', 'comedy', 'drama', 'thriller', 'children', 'adventure'}
### Row template looks like --> {'title': 0, 'horror': 0, 'comedy': 0, 'drama': 0, 'children': 0, 'mystery': 0, 'movieId': 0, 'thriller': 0, 'adventure': 0}
### Row looks like {'title': 'Big Night (1996)', 'horror': 0, 'comedy': 1, 'drama': 1, 'children': 0, 'mystery': 0, 'movieId': 994, 'thriller': 0, 'adventure': 0}
### Row looks like {'title': 'Grudge, The (2004)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 0, 'mystery': 1, 'movieId': 8947, 'thriller': 1, 'adventure': 0}
### Row looks like {'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}
### Collection looks like [{'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}, {'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}, {'title': 'Cheetah (1989)', 'horror': 1, 'comedy': 1, 'drama': 1, 'children': 1, 'mystery': 1, 'movieId': 2039, 'thriller': 1, 'adventure': 1}]
I want my collection to like
### [
{'title': 'Big Night (1996)', 'horror': 0, 'comedy': 1, 'drama': 1, 'children': 0, 'mystery': 0, 'movieId': 994, 'thriller': 0, 'adventure': 0},
{'title': 'Grudge, The (2004)', 'horror': 1, 'comedy': 0, 'drama': 0, 'children': 0, 'mystery': 1, 'movieId': 8947, 'thriller': 1, 'adventure': 0},
{'title': 'Cheetah (1989)', 'horror': 0, 'comedy': 0, 'drama': 0, 'children': 1, 'mystery': 0, 'movieId': 2039, 'thriller': 0, 'adventure': 1}
]
Dataset - https://grouplens.org/datasets/movielens/


I got to understand the issue now, i was trying to mutate the dictionary object.
def tranformDataFrame(self):
genres = self.__findUniqueGenres();
print('### List of genres...', genres);
__df = self.__prepareDataframe(genres); # Data frame with all required columns.
rowTemplate = self.__getRowTemplate(__df.columns)
print('### Row template looks like -->', rowTemplate)
collection = []
for index, row in self.df.iterrows():
# Creating the fresh copy of row template every time prevent mutation.
_rowToAdd = self.__createRow(self.__getRowTemplate(__df.columns), row);
print('### Row looks like', _rowToAdd)
collection.append(_rowToAdd)
print('### Collection looks like', collection)
return __df.append(collection)
Although there must be some way to cache the copy and cloning it every time ( instead of processing some logic, and creating a dictionary). But, this solution resolve this particular issue at-least.

Related

sorting a nested dictionary by 3 criteria

I have a nested dictionary and I want to sort it by 3 fields, firstly by points secondly by wins(if points are equal) and lastly by alphabet if both points and wins are equal.
My dictionary:
{'Iran': {'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1},
'Morocco': {'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1},
'Portugal': {'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1},
'Spain': {'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1}}
My code:
sort_data=sorted(dic.keys(),key=lambda x:(-dic[x]["points"],dic[x]["wins"]))
for items in sort_data:
x=dic[items]
print(items," ",str(x).replace("'", "").replace("{","").replace("}", ""))
my code dose not work in this case and dose not sort it by alphabet when all the situations are equal.Could you help me?

I suggest using dic.items() rather than dic.keys(), although that's mostly a matter of taste.
def sorted_dict(dic):
def key(x):
k,v = x
return (v['points'], v['wins'], k)
return dict(sorted(dic.items(), key=key))
dic1 = {'Iran': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1}, 'Morocco': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1}, 'Portugal': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1}, 'Spain': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1}}
dic2 = sorted_dict(dic1)
print(dic2)
# {'Iran': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1},
# 'Morocco': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1},
# 'Portugal': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1},
# 'Spain': {'draws': 1, 'goal difference': 0, 'loses': 1, 'points': 4, 'wins': 1}}
Note that in your dictionary, all countries have the same number of points and wins, and the countries are already sorted in alphabetical order, so this is a terrible example to test whether the sort worked correctly or not.

You could use dict.items with a lambda instead:
import json # For pretty printing only.
table = {
'Iran': {
'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1
},
'Morocco': {
'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1
},
'Portugal': {
'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1
},
'Spain': {
'draws': 1,
'goal difference': 0,
'loses': 1,
'points': 4,
'wins': 1
}
}
sorted_table = dict(
sorted(table.items(),
key=lambda kvp: (-kvp[1]['points'], -kvp[1]['wins'], kvp[0])))
print(json.dumps(sorted_table, indent=4))
Output:
{
"Iran": {
"draws": 1,
"goal difference": 0,
"loses": 1,
"points": 4,
"wins": 1
},
"Morocco": {
"draws": 1,
"goal difference": 0,
"loses": 1,
"points": 4,
"wins": 1
},
"Portugal": {
"draws": 1,
"goal difference": 0,
"loses": 1,
"points": 4,
"wins": 1
},
"Spain": {
"draws": 1,
"goal difference": 0,
"loses": 1,
"points": 4,
"wins": 1
}
}
Note: No change from input is because input is already sorted...

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}

Cummulative Dictionary

I am trying to write a python function where for each key (the dates), the value would be the sum of that day's result and the previous day(s) (sort of following the same logic as the fibonacci sequence).
For example, I have:
{20200516: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 0}, 20200517: {'Level1': 0, 'Level2': 0, 'Level3': 0, 'Level4': 1}, 20200518: {'Level1': 1, 'Level2': 0, 'Level3': 0, 'Level4': 0}, 20200519: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 1}}
but I want to have:
{20200516: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 0}, 20200517: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 1}, 20200518: {'Level1': 1, 'Level2': 1, 'Level3': 0, 'Level4': 1}, 20200519: {'Level1': 1, 'Level2': 2, 'Level3': 0, 'Level4': 2}
What I have done until now:
def summing(d):
'''
each key after the first one is the sum of the one before and its own result
>>> {20200516: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 0}, 20200517: {'Level1': 0,
'Level2': 0, 'Level3': 0, 'Level4': 1}, 20200518: {'Level1': 1, 'Level2': 0, 'Level3':
0, 'Level4': 0}, 20200519: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 1}}
{20200516: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 0}, 20200517: {'Level1': 0,
'Level2': 1, 'Level3': 0, 'Level4': 1}, 20200518: {'Level1': 1, 'Level2': 1, 'Level3': 0, '
Level4': 1}, 20200519: {'Level1': 1, 'Level2': 2, 'Level3': 0, 'Level4': 2}
'''
#STILL IN PROGRESS
c={}
for key in d:
if key == 20200516:
c[20200516]=d[20200516]
else:
c[key]=d[key-1]+d[key]
return c

You made a good effort, but you can't just add dicts like that. Here's a minimal change to get from your input to desired output, by using dict comprehension to add the value for each entry in the daily record:
from pprint import pprint
def summing_oneday(d1, d2):
return {key: d1[key] + d2[key] for key in d2}
def summing(data):
result = {}
for day in sorted(data.keys()):
if not result:
result[day] = data[day]
else:
result[day] = summing_oneday(previous, data[day])
previous = result[day]
return result
data = {20200516: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 0}, 20200517: {'Level1': 0, 'Level2': 0, 'Level3': 0, 'Level4': 1}, 20200518: {'Level1': 1, 'Level2': 0, 'Level3': 0, 'Level4': 0}, 20200519: {'Level1': 0, 'Level2': 1, 'Level3': 0, 'Level4': 1}}
pprint(summing(data))
I'm assuming all the keys are present on all the daily records. Otherwise we'll have to deal with that.

MongoDB - use numericOrdering in aggregation do mongo slow

I have this issue, when I run my query normal, everything works 100% perfect but when I try to add numericOrdering it will not.
aggregation = [
{'$match': {'store': 'fourcom', 'closed': 0}},
{'$lookup': {
'from': 'ordre_open',
'localField': 'reference_number',
'foreignField': 'order_number',
'as': 'ordre_open'
}},
{'$lookup': {
'from': 'ordre_backend_open',
'localField': 'reference_number',
'foreignField': 'ordre-id',
'as': 'ordre_backend'
}},
{'$project': {
'_id': 1,
'store': 1,
'order_number': 1,
'document_number': 1,
'reference_number': 1,
'document_date': 1,
'invoice_group': 1,
'account': 1,
'name': 1,
'our_ref': 1,
'your_ref': 1,
'payment': 1,
'vat': 1,
'total_price': 1,
'currency': 1,
'department': 1,
'closed': 1,
'deleted': 1,
'arch': 1,
'locations': 1,
'lines.price': 1,
'lines.qty': 1,
'ordre_open._id': 1,
'ordre_open.order_number': 1,
'ordre_open.closed': 1,
'ordre_open.deleted': 1,
'ordre_open.type': 1,
'ordre_open.arch': 1,
'ordre_open.fcomputer_synced': 1,
'ordre_backend.ordre-id': 1,
'ordre_backend.error': 1,
'ordre_backend.done': 1,
'ordre_backend.cancel': 1
}},
{'$sort': {'order_number': 1}},
{'$skip': 0},
{'$limit': 1000}
]
mongodb_limit = 1000
resualt = db.command('aggregate', 'ordre_purchase_open', pipeline=aggregation, explain=False, cursor={
'batchSize': mongodb_limit
})
but when I decide to to sort like we humans do and add
collation={
'numericOrdering': True,
'locale': "da"
}
its going total wrong, my query go down and will not work as I expect. I have used numericOrdering many time before but this time is first time in db.command by using aggregate.
I hope someone can explain what I am doing wrong here?

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':{}})

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