Pandas help!
I have a specific column like this,
Mpg
0 18
1 17
2 19
3 21
4 16
5 15
Mpg is mile per gallon,
Now I need to replace that 'MPG' column to 'litre per 100 km' and change those values to litre per 100 km' at the same time. Any help? Thanks beforehand.
-Tom
I changed the name of the column but doing both simultaneously,i could not.
Use pop to return and delete the column at the same time and rdiv to perform the conversion (1 mpg = 1/235.15 liter/100km):
df['litre per 100 km'] = df.pop('Mpg').rdiv(235.15)
If you want to insert the column in the same position:
df.insert(df.columns.get_loc('Mpg'), 'litre per 100 km',
df.pop('Mpg').rdiv(235.15))
Output:
litre per 100 km
0 13.063889
1 13.832353
2 12.376316
3 11.197619
4 14.696875
5 15.676667
An alternative to pop would be to store the result in another dataframe. This way you can perform the two steps at the same time. In my code below, I first reproduce your dataframe, then store the constant for conversion and perform it on all entries using the apply method.
df = pd.DataFrame({'Mpg':[18,17,19,21,16,15]})
cc = 235.214583 # constant for conversion from mpg to L/100km
df2 = pd.DataFrame()
df2['litre per 100 km'] = df['Mpg'].apply(lambda x: cc/x)
print(df2)
The output of this code is:
litre per 100 km
0 13.067477
1 13.836152
2 12.379715
3 11.200694
4 14.700911
5 15.680972
as expected.
Related
I have a df containing sub-trajectories (segments) of users, with mode of travel indicated by 0,1,2... which looks like this:
df = pd.read_csv('sample.csv')
df
id lat lon mode
0 5138001 41.144540 -8.562926 0
1 5138001 41.144538 -8.562917 0
2 5138001 41.143689 -8.563012 0
3 5138003 43.131562 -8.601273 1
4 5138003 43.132107 -8.598124 1
5 5145001 37.092095 -8.205070 0
6 5145001 37.092180 -8.204872 0
7 5145015 39.289341 -8.023454 2
8 5145015 39.197432 -8.532761 2
9 5145015 39.198361 -8.375641 2
In the above sample, id is for the segments but a full trajectory maybe covered by different modes (i.e. contains multiple segments).
So the first 4-digits of id is the unique trajectories, and the last 3-digits, unique segment with that trajectory.
I know that I can count the number of unique segments in the dfusing:
df.groupby('id').['mode'].nunique()
How do I then count the number of unique trajectories 5138, 5145, ...?
Use indexing for get first 4 values with str, if necessary first convert values to strings by Series.astype:
df = df.groupby(df['id'].astype(str).str[:4])['mode'].nunique().reset_index(name='count')
print (df)
id count
0 5138 2
1 5145 2
If need processing values after first 4 ids:
s = df['id'].astype(str)
df = s.str[4:].groupby(s.str[:4]).nunique().reset_index(name='count')
print (df)
id count
0 5138 2
1 5145 2
Another idea is use lambda function:
df.groupby(df['id'].apply(lambda x: str(x)[:4]))['mode'].nunique()
Is there a way to use numpy to add numbers in a series up to a threshold, then restart the counter. The intention is to form groupby based on the categories created.
amount price
0 27 22.372505
1 17 126.562276
2 33 101.061767
3 78 152.076373
4 15 103.482099
5 96 41.662766
6 108 98.460743
7 143 126.125865
8 82 87.749286
9 70 56.065133
The only solutions I found iterate with .loc which is slow. I tried building a solution based on this answer https://stackoverflow.com/a/56904899:
sumvals = np.frompyfunc(lambda a,b: a+b if a <= 100 else b,2,1)
df['cumvals'] = sumvals.accumulate(df['amount'], dtype=np.object)
The use-case is to find the average price of every 75 sold amounts of the thing.
Solution #1 Interpreting the following one way will get my solution below: "The use-case is to find the average price of every 75 sold amounts of the thing." If you are trying to do this calculation the "hard way" instead of pd.cut, then here is a solution that will work well but the speed / memory will depend on the cumsum() of the amount column, which you can find out if you do df['amount'].cumsum(). The output will take about 1 second per every 10 million of the cumsum, as that is how many rows is created with np.repeat. Again, this solution is not horrible if you have less than ~10 million in cumsum (1 second) or even 100 million in cumsum (~10 seconds):
i = 75
df = np.repeat(df['price'], df['amount']).to_frame().reset_index(drop=True)
g = df.index // i
df = df.groupby(g)['price'].mean()
df.index = (df.index * i).astype(str) + '-' + (df.index * i +75).astype(str)
df
Out[1]:
0-75 78.513748
75-150 150.715984
150-225 61.387540
225-300 67.411182
300-375 98.829611
375-450 126.125865
450-525 122.032363
525-600 87.326831
600-675 56.065133
Name: price, dtype: float64
Solution #2 (I believe this is wrong but keeping just in case)
I do not believe you are tying to do it this way, which was my initial solution, but I will keep it here in case, as you haven't included expected output. You can create a new series with cumsum and then use pd.cut and pass bins=np.arange(0, df['Group'].max(), 75) to create groups of cumulative 75. Then, groupby the groups of cumulative 75 and take the mean. Finally, use pd.IntervalIndex to clean up the format and change to a sting:
df['Group'] = df['amount'].cumsum()
s = pd.cut(df['Group'], bins=np.arange(0, df['Group'].max(), 75))
df = df.groupby(s)['price'].mean().reset_index()
df['Group'] = pd.IntervalIndex(df['Group']).left.astype(str) + '-' + pd.IntervalIndex(df['Group']).right.astype(str)
df
Out[1]:
Group price
0 0-75 74.467390
1 75-150 101.061767
2 150-225 127.779236
3 225-300 41.662766
4 300-375 98.460743
5 375-450 NaN
6 450-525 126.125865
7 525-600 87.749286
My code uses a column called booking status that is 1 for yes and 0 for no (there are multiple other columns that information will be pulled from dependant on the booking status) - there are lots more no than yes so I would like to take a sample with all the yes and the same amount of no.
When I use
samp = rslt_df.sample(n=298, random_state=1, weights='bookingstatus')
I get the error:
ValueError: Fewer non-zero entries in p than size
Is there a way to do this sample this way?
If our entire dataset looks like this:
print(df)
c1 c2
0 1 1
1 0 2
2 0 3
3 0 4
4 0 5
5 0 6
6 0 7
7 1 8
8 0 9
9 0 10
We may decide to sample from it using the DataFrame.sample function. By default, this function will sample without replacement. Meaning, you'll receive an error by specifying a number of observations larger than the number of observations in your initial dataset:
df.sample(20)
ValueError: Cannot take a larger sample than population when 'replace=False'
In your situation, the ValueError comes from the weights parameter:
df.sample(3,weights='c1')
ValueError: Fewer non-zero entries in p than size
To paraphrase the DataFrame.sample docs, using the c1 column as our weights parameter implies that rows with a larger value in the c1 column are more likely to be sampled. Specifically, the sample function will not pick values from this column that are zero. We can fix this error using either one of the following methods.
Method 1: Set the replace parameter to be true:
m1 = df.sample(3,weights='c1', replace=True)
print(m1)
c1 c2
0 1 1
7 1 8
0 1 1
Method 2: Make sure the n parameter is equal to or less than the number of 1s in the c1 column:
m2 = df.sample(2,weights='c1')
print(m2)
c1 c2
7 1 8
0 1 1
If you decide to use this method, you won't really be sampling. You're really just filtering out any rows where the value of c1 is 0.
I was able to this in the end, here is how I did it:
bookingstatus_count = df.bookingstatus.value_counts()
print('Class 0:', bookingstatus_count[0])
print('Class 1:', bookingstatus_count[1])
print('Proportion:', round(bookingstatus_count[0] / bookingstatus_count[1], 2), ': 1')
# Class count
count_class_0, count_class_1 = df.bookingstatus.value_counts()
# Divide by class
df_class_0 = df[df['bookingstatus'] == 0]
df_class_0_under = df_class_0.sample(count_class_1)
df_test_under = pd.concat([f_class_0_under, df_class_1], axis=0)
df_class_1 = df[df['bookingstatus'] == 1]
based on this https://www.kaggle.com/rafjaa/resampling-strategies-for-imbalanced-datasets
Thanks everyone
A dummy dataset is :
data <- data.frame(
group = c(1,1,1,1,1,2),
dates = as.Date(c("2005-01-01", "2006-05-01", "2007-05-01","2004-08-01",
"2005-03-01","2010-02-01")),
value = c(10,20,NA,40,NA,5)
)
For each group, the missing values need to be filled with the non-missing value corresponding to the nearest date within same group. In case of a tie, pick any.
I am using dplyr. which.closest from birk but it needs a vector and a value. How to look up within a vector without writing loops. Even if there is an SQL solution, will do.
Any pointers to the solution?
May be something like: value = value[match(which.closest(dates,THISdate) & !is.na(value))]
Not sure how to specify Thisdate.
Edit: The expected value vector should look like:
value = c(10,20,20,40,10,5)
Using knn1 (nearest neighbor) from the class package (which comes with R -- don't need to install it) and dplyr define an na.knn1 function which replaces each NA value in x with the non-NA x value having the closest time.
library(class)
na.knn1 <- function(x, time) {
is_na <- is.na(x)
if (sum(is_na) == 0 || all(is_na)) return(x)
train <- matrix(time[!is_na])
test <- matrix(time[is_na])
cl <- x[!is_na]
x[is_na] <- as.numeric(as.character(knn1(train, test, cl)))
x
}
data %>% mutate(value = na.knn1(value, dates))
giving:
group dates value
1 1 2005-01-01 10
2 1 2006-05-01 20
3 1 2007-05-01 20
4 1 2004-08-01 40
5 1 2005-03-01 10
6 2 2010-02-01 5
Add an appropriate group_by if the intention was to do this by group.
You can try the use of sapply to find the values closest since the x argument in `which.closest only takes a single value.
first create a vect whereby the dates with no values are replaced with NA and use it within the which.closest function.
library(birk)
vect=replace(data$dates,which(is.na(data$value)),NA)
transform(data,value=value[sapply(dates,which.closest,vec=vect)])
group dates value
1 1 2005-01-01 10
2 1 2006-05-01 20
3 1 2007-05-01 20
4 1 2004-08-01 40
5 1 2005-03-01 10
6 2 2010-02-01 5
if which.closest was to take a vector then there would be no need of sapply. But this is not the case.
Using the dplyr package:
library(birk)
library(dplyr)
data%>%mutate(vect=`is.na<-`(dates,is.na(value)),
value=value[sapply(dates,which.closest,vect)])%>%
select(-vect)
The rolling function in Pandas can only calculate rolling statistics according to row counts or date/time columns. But I want to have a discrete time column for calculating rolling sum, something like this:
key time value
A 1 10
A 2 20
A 4 30
A 7 10
B 1 15
B 2 30
B 3 15
I want to first group by key, then calculate the rolling sum on value for the nearest 3 time:
key time value output
A 1 10 10
A 2 20 30(10+20)
A 4 30 60(10+20+30)
A 7 10 40(30+10)
B 1 15 15
B 2 30 45
B 3 15 60
I tried this:
grouped = input.groupby("key", as_index=False)
for name, group in grouped:
group = group.sort_values("time")
time = list(group["time"])
value = list(group["value"])
#calcRollingStat is a custom function that outputs a list of corresponding results
out = calcRollingStat(time, value, mode="avg")
group["output"] = out #out is a list
But then I don't know how to convert grouped back to DataFrame. Pandas tells me that there is no reset_index attribute in grouped.
Is my code the best method to do this? How would you tackle this problem?
Thank you!
I believe you can use GroupBy.apply with custom function:
def f(group):
group = group.sort_values("time")
time = list(group["time"])
value = list(group["value"])
#calcRollingStat is a custom function that outputs a list of corresponding results
group["output"] = calcRollingStat(time, value, mode="avg")
return group
df = input.groupby("key", as_index=False).apply(f)