I am trying to calculate the biggest difference between summer gold medal counts and winter gold medal counts relative to their total gold medal count. The problem is that I need to consider only countries that have won at least 1 gold medal in both summer and winter.
Gold: Count of summer gold medals
Gold.1: Count of winter gold medals
Gold.2: Total Gold
This a sample of my data:
Gold Gold.1 Gold.2 ID diff gold %
Afghanistan 0 0 0 AFG NaN
Algeria 5 0 5 ALG 1.000000
Argentina 18 0 18 ARG 1.000000
Armenia 1 0 1 ARM 1.000000
Australasia 3 0 3 ANZ 1.000000
Australia 139 5 144 AUS 0.930556
Austria 18 59 77 AUT 0.532468
Azerbaijan 6 0 6 AZE 1.000000
Bahamas 5 0 5 BAH 1.000000
Bahrain 0 0 0 BRN NaN
Barbados 0 0 0 BAR NaN
Belarus 12 6 18 BLR 0.333333
This is the code that I have but it is giving the wrong answer:
def answer():
Gold_Y = df2[(df2['Gold'] > 1) | (df2['Gold.1'] > 1)]
df2['difference'] = (df2['Gold']-df2['Gold.1']).abs()/df2['Gold.2']
return df2['diff gold %'].idxmax()
answer()
Try this code after subbing in the correct (your) function and variable names. I'm new to Python, but I think the issue was that you had to use the same variable in Line 4 (df1['difference']), and just add the method (.idxmax()) to the end. I don't think you need the first line of code for the function, either, as you don't use the local variable (Gold_Y). FYI - I don't think we're working with the same dataset.
def answer_three():
df1['difference'] = (df1['Gold']-df1['Gold.1']).abs()/df1['Gold.2']
return df1['difference'].idxmax()
answer_three()
def answer_three():
atleast_one_gold = df[(df['Gold']>1) & (df['Gold.1']> 1)]
return ((atleast_one_gold['Gold'] - atleast_one_gold['Gold.1'])/atleast_one_gold['Gold.2']).idxmax()
answer_three()
def answer_three():
_df = df[(df['Gold'] > 0) & (df['Gold.1'] > 0)]
return ((_df['Gold'] - _df['Gold.1']) / _df['Gold.2']).argmax() answer_three()
This looks like a question from the programming assignment of courser course -
"Introduction to Data Science in Python"
Having said that if you are not cheating "maybe" the bug is here:
Gold_Y = df2[(df2['Gold'] > 1) | (df2['Gold.1'] > 1)]
You should use the & operator. The | operator means you have countries that have won Gold in either the Summer or Winter olympics.
You should not get a NaN in your diff gold.
def answer_three():
diff=df['Gold']-df['Gold.1']
relativegold = diff.abs()/df['Gold.2']
df['relativegold']=relativegold
x = df[(df['Gold.1']>0) &(df['Gold']>0) ]
return x['relativegold'].idxmax(axis=0)
answer_three()
I an pretty new to python or programming as a whole.
So my solution would be the most novice ever!
I love to create variables; so you'll see a lot in the solution.
def answer_three:
a = df.loc[df['Gold'] > 0,'Gold']
#Boolean masking that only prints the value of Gold that matches the condition as stated in the question; in this case countries who had at least one Gold medal in the summer seasons olympics.
b = df.loc[df['Gold.1'] > 0, 'Gold.1']
#Same comment as above but 'Gold.1' is Gold medals in the winter seasons
dif = abs(a-b)
#returns the abs value of the difference between a and b.
dif.dropna()
#drops all 'Nan' values in the column.
tots = a + b
#i only realised that this step wasn't essential because the data frame had already summed it up in the column 'Gold.2'
tots.dropna()
result = dif.dropna()/tots.dropna()
returns result.idxmax
# returns the index value of the max result
def answer_two():
df2=pd.Series.max(df['Gold']-df['Gold.1'])
df2=df[df['Gold']-df['Gold.1']==df2]
return df2.index[0]
answer_two()
def answer_three():
return ((df[(df['Gold']>0) & (df['Gold.1']>0 )]['Gold'] - df[(df['Gold']>0) & (df['Gold.1']>0 )]['Gold.1'])/df[(df['Gold']>0) & (df['Gold.1']>0 )]['Gold.2']).argmax()
Related
I have been stuck to a problem where I have done all the groupby operation and got the resultant dataframe as shown below but the problem came in last operation of calculation of one additional column
Current dataframe:
code industry category count duration
2 Retail Mobile 4 7
3 Retail Tab 2 33
3 Health Mobile 5 103
2 Food TV 1 88
The question: Want an additional column operation which calculates the ratio of count of industry 'retail' for the specific code column entry
for example: code 2 has 2 industry entry retail and food so operation column should have value 4/(4+1) = 0.8 and similarly for code3 as well as shown below
O/P:
code industry category count duration operation
2 Retail Mobile 4 7 0.8
3 Retail Tab 2 33 -
3 Health Mobile 5 103 2/7 = 0.285
2 Food TV 1 88 -
Help on here as well that if I do just groupby I will miss out the information of category and duration also what would be better way to represent the output df there can been multiple industry and operation is limited to just retail
I can't think of a single operation. But the way via a dictionary should work. Oh, and in advance for the other answerers the code to create the example dataframe.
st_l = [[2,'Retail','Mobile', 4, 7],
[3,'Retail', 'Tab', 2, 33],
[3,'Health', 'Mobile', 5, 103],
[2,'Food', 'TV', 1, 88]]
df = pd.DataFrame(st_l, columns=
['code','industry','category','count','duration'])
And now my attempt:
sums = df[['code', 'count']].groupby('code').sum().to_dict()['count']
df['operation'] = df.apply(lambda x: x['count']/sums[x['code']], axis=1)
You can create a new column with the total count of each code using groupby.transform(), and then use loc to find only the rows that have as their industry 'Retail' and perform your division:
df['total_per_code'] = df.groupby(['code'])['count'].transform('sum')
df.loc[df.industry.eq('Retail'), 'operation'] = df['count'].div(df.total_per_code)
df.drop('total_per_code',axis=1,inplace=True)
prints back:
code industry category count duration operation
0 2 Retail Mobile 4 7 0.800000
1 3 Retail Tab 2 33 0.285714
2 3 Health Mobile 5 103 NaN
3 2 Food TV 1 88 NaN
I need to search the values from the df1['numsearch'] column into the lists in df2['Numbers']. If the number is in those lists, then I want to add values from the df2['Score'] column to df1. See desired output below.
df1 = pd.DataFrame(
{'Day':['M','Tu','W','Th','Fr','Sa','Su'],
'numsearch':['1','20','14','99','19','6','101']
})
df2 = pd.DataFrame(
{'Letters':['a','b','c','d'],
'Numbers':[['1','2','3','4'],['5','6','7','8'],['10','20','30','40'],['11','12','13','14']],
'Score': ['1.1','2.2','3.3','4.4']})
desired output
Day numsearch Score
0 M 1 1.1
1 Tu 20 3.3
2 W 4 4.4
3 Th 99 "No score"
4 Fr 19 "No score"
5 Sa 6 2.2
6 Su 101 "No score"
I have written a for loop that works with the test data.
scores = []
for s,ns in enumerate(ppr_data['SN']):
match = ''
for k,q in enumerate(jcr_data['All_ISSNs']):
if ns in q:
scores.append(jcr_data['Journal Impact Factor'][k])
match = 1
else:
continue
if match == "":
scores.append('No score')
match = ""
df1['Score'] = np.array(scores)
In my small test, but above code works, but when working with larger data files, it is creating duplicates. So this clearly isn't the best way to do this.
I'm sure there's a more pandas-proper line of code that ends in .fillna("No score") .
I tried to use a loc statement, but I get hung up on searching the values of one dataframe in a column that contains lists.
Can anyone shed some light?
df2=df2.explode('Numbers')#Explode df2 on Numbers
d=dict(zip(df2.Numbers, df2.Score))#dict Numbers and Scores
df1['Score']=df1.numsearch.map(d).fillna('No Score')#Map dict to df1 filling NaN with No Score
Can shorten it as follows:
df2=df2.explode('Numbers')#Explode df2 on Numbers
df1['Score']=df1.numsearch.map(dict(zip(df2.Numbers, df2.Score))).fillna('No Score')
Day numsearch Score
0 M 1 1.1
1 Tu 20 3.3
2 W 14 4.4
3 Th 99 No Score
4 Fr 19 No Score
5 Sa 6 2.2
6 Su 101 No Score
You can try left join and fillna:
df1.merge(df2.explode('Numbers'),
left_on='numsearch',
right_on='Numbers', how='left')[['Day', 'numsearch', 'Score']].fillna("No score")
Output:
Day numsearch Score
0 M 1 1.1
1 Tu 20 3.3
2 W 14 4.4
3 Th 99 No score
4 Fr 19 No score
5 Sa 6 2.2
6 Su 101 No score
I have a pandas dataframe as follows:
Name Age City Country percentage
a Jack 34 Sydney Australia 0.23
b Riti 30 Delhi India 0.45
c Vikas 31 Mumbai India 0.55
d Neelu 32 Bangalore India 0.73
e John 16 New York US 0.91
f Mike 17 las vegas US 0.78
I am planning to add one more column called bucket whose definition depends on the percentage column as follows:
less than 0.25 = 1
between 0.25 and 0.5 = 2
between 0.5 and 0.75 = 3
greater than 0.75 = 4
I tried the inbuilt conditions and choices properties of pandas follows:
conditions = [(df_obj['percentage'] < .25),
(df_obj['percentage'] >=.25 & df_obj['percentage'] < .5),
(df_obj['percentage'] >=.5 & df_obj['percentage'] < .75),
(df_obj['percentage'] >= .75)]
choices = [1,2,3,4]
df_obj['bucket'] = np.select(conditions, choices)
However, this gives me a random error as follows in the line where I create the conditions:
TypeError: Cannot perform 'rand_' with a dtyped [float64] array and scalar of type [bool]
A quick fix to your code is that you need more parentheses, for example:
((df_obj['percentage'] >=.25) & (df_obj['percentage'] < .5) )
^ ^ ^ ^
However, I think it's cleaner with pd.cut:
pd.cut(df['percentage'], bins=[0,0.25, 0.5, 0.75, 1],
include_lowest=True, right=False,
labels=[1,2,3,4])
Or since your buckets are linear:
df['bucket'] = (df['percentage']//0.25).add(1).astype(int)
Output
Name Age City Country percentage bucket
a Jack 34 Sydney Australia 0.23 1
b Riti 30 Delhi India 0.45 2
c Vikas 31 Mumbai India 0.55 3
d Neelu 32 Bangalore India 0.73 3
e John 16 New York US 0.91 4
f Mike 17 las vegas US 0.78 4
I think the easiest/most readable way to do this is to use the apply function:
def percentage_to_bucket(percentage):
if percentage < .25:
return 1
elif percentage >= .25 and percentage < .5:
return 2
elif percentage >= .5 and percentage < .75:
return 3
else:
return 4
df["bucket"] = df["percentage"].apply(percentage_to_bucket)
Pandas apply will take each value of a given column and apply the passed function to this value, returning a pandas series with the results, which you can then assign to your new column.
For the given dataframe as follows:
id|address|sell_price|market_price|status|start_date|end_date
1|7552 Atlantic Lane|1170787.3|1463484.12|finished|2019/8/2|2019/10/1
1|7552 Atlantic Lane|1137782.02|1422227.52|finished|2019/8/2|2019/10/1
2|888 Foster Street|1066708.28|1333385.35|finished|2019/8/2|2019/10/1
2|888 Foster Street|1871757.05|1416757.05|finished|2019/10/14|2019/10/15
2|888 Foster Street|NaN|763744.52|current|2019/10/12|2019/10/13
3|5 Pawnee Avenue|NaN|928366.2|current|2019/10/10|2019/10/11
3|5 Pawnee Avenue|NaN|2025924.16|current|2019/10/10|2019/10/11
3|5 Pawnee Avenue|Nan|4000000|forward|2019/10/9|2019/10/10
3|5 Pawnee Avenue|2236138.9|1788938.9|finished|2019/10/8|2019/10/9
4|916 W. Mill Pond St.|2811026.73|1992026.73|finished|2019/9/30|2019/10/1
4|916 W. Mill Pond St.|13664803.02|10914803.02|finished|2019/9/30|2019/10/1
4|916 W. Mill Pond St.|3234636.64|1956636.64|finished|2019/9/30|2019/10/1
5|68 Henry Drive|2699959.92|NaN|failed|2019/10/8|2019/10/9
5|68 Henry Drive|5830725.66|NaN|failed|2019/10/8|2019/10/9
5|68 Henry Drive|2668401.36|1903401.36|finished|2019/12/8|2019/12/9
#copy above data and run below code to reproduce dataframe
df = pd.read_clipboard(sep='|')
I would like to groupby id and address and calculate mean_ratio and result_count based on the following conditions:
mean_ratio: which is groupby id and address and calculate mean for the rows meet the following conditions: status is finished and start_date isin the range of 2019-09 and 2019-10
result_count: which is groupby id and address and count the rows meet the following conditions: status is either finished or failed, and start_date isin the range of 2019-09 and 2019-10
The desired output will like this:
id address mean_ratio result_count
0 1 7552 Atlantic Lane NaN 0
1 2 888 Foster Street 1.32 1
2 3 5 Pawnee Avenue 1.25 1
3 4 916 W. Mill Pond St. 1.44 3
4 5 68 Henry Drive NaN 2
I have tried so far:
# convert date
df[['start_date', 'end_date']] = df[['start_date', 'end_date']].apply(lambda x: pd.to_datetime(x, format = '%Y/%m/%d'))
# calculate ratio
df['ratio'] = round(df['sell_price']/df['market_price'], 2)
In order to filter start_date isin the range of 2019-09 and 2019-10:
L = [pd.Period('2019-09'), pd.Period('2019-10')]
c = ['start_date']
df = df[np.logical_or.reduce([df[x].dt.to_period('m').isin(L) for x in c])]
To filter row status is finished or failed, I use:
mask = df['status'].str.contains('finished|failed')
df[mask]
But I don't know how to use those to get final result. Thanks your help at advance.
I think you need GroupBy.agg, but because some rows are excluded like id=1, then add them by DataFrame.join with all unique pairs id and address in df2, last replace missing values in result_count columns:
df2 = df[['id','address']].drop_duplicates()
print (df2)
id address
0 1 7552 Atlantic Lane
2 2 888 Foster Street
5 3 5 Pawnee Avenue
9 4 916 W. Mill Pond St.
12 5 68 Henry Drive
df[['start_date', 'end_date']] = df[['start_date', 'end_date']].apply(lambda x: pd.to_datetime(x, format = '%Y/%m/%d'))
df['ratio'] = round(df['sell_price']/df['market_price'], 2)
L = [pd.Period('2019-09'), pd.Period('2019-10')]
c = ['start_date']
mask = df['status'].str.contains('finished|failed')
mask1 = np.logical_or.reduce([df[x].dt.to_period('m').isin(L) for x in c])
df = df[mask1 & mask]
df1 = df.groupby(['id', 'address']).agg(mean_ratio=('ratio','mean'),
result_count=('ratio','size'))
df1 = df2.join(df1, on=['id','address']).fillna({'result_count': 0})
print (df1)
id address mean_ratio result_count
0 1 7552 Atlantic Lane NaN 0.0
2 2 888 Foster Street 1.320000 1.0
5 3 5 Pawnee Avenue 1.250000 1.0
9 4 916 W. Mill Pond St. 1.436667 3.0
12 5 68 Henry Drive NaN 2.0
Some helpers
def mean_ratio(idf):
# filtering data
idf = idf[
(idf['start_date'].between('2019-09-01', '2019-10-31')) &
(idf['mean_ratio'].notnull()) ]
return np.round(idf['mean_ratio'].mean(), 2)
def result_count(idf):
idf = idf[
(idf['status'].isin(['finished', 'failed'])) &
(idf['start_date'].between('2019-09-01', '2019-10-31')) ]
return idf.shape[0]
# We can caluclate `mean_ratio` before hand
df['mean_ratio'] = df['sell_price'] / df['market_price']
df = df.astype({'start_date': np.datetime64, 'end_date': np.datetime64})
# Group the df
g = df.groupby(['id', 'address'])
mean_ratio = g.apply(lambda idf: mean_ratio(idf)).to_frame('mean_ratio')
result_count = g.apply(lambda idf: result_count(idf)).to_frame('result_count')
# Final result
pd.concat((mean_ratio, result_count), axis=1)
I'm trying to convert wiki page table to dataframe. Headings are shifted to the
right, 'Launches' should be there were it is now 'Successes'.
I have used skiprows option, but it did not work.
df = pd.read_html(r'https://en.wikipedia.org/wiki/2018_in_spaceflight',skiprows=[1,2])[7]
df2 = df[df.columns[1:5]]
1 2 3 4
0 Launches Successes Failures Partial failures
1 India 1 1 0
2 Japan 3 3 0
3 New Zealand 1 1 0
4 Russia 3 3 0
5 United States 8 8 0
6 24 23 0 1
The problem is there are merged cells in the first column of the original table. If you want to parse it exactly, you should write a parser. Provisionally, you can try:
df = pd.read_html(r'https://en.wikipedia.org/wiki/2018_in_spaceflight', header=0)[7]
df.columns = [""] + list(df.columns[:-1])
df.iloc[-1] = [""] + list(df.iloc[-1][:-1])