I am new to pandas. I know how to use drop_duplicates and take the last observed row in a dataframe. Is there any way that I can use it to take only second last observed. Or any other way of doing it.
For example:
I would like to go from
df = pd.DataFrame(data={'A':[1,1,1,2,2,2],'B':[1,2,3,4,5,6]}) to
df1 = pd.DataFrame(data={'A':[1,2],'B':[2,5]})
The idea is that you'll group the data by the duplicate column , then check the length of group , if the length of group is greater than or equal 2 this mean that you can slice the second element of group , if the group has a length of one which mean that this value is not duplicated , then take index 0 which is the only element in the grouped data
df.groupby(df['A']).apply(lambda x : x.iloc[1] if len(x) >= 2 else x.iloc[0])
The first answer I think was on the right track, but possibly not quite right. I have extended your data to include 'A' groups with two observations, and an 'A' group with one observation, for the sake of completeness.
import pandas as pd
df = pd.DataFrame(data={'A':[1,1,1,2,2,2, 3, 3, 4],'B':[1,2,3,4,5,6, 7, 8, 9]})
def user_apply_func(x):
if len(x) == 2:
return x.iloc[0]
if len(x) > 2:
return x.iloc[-2]
return
df.groupby('A').apply(user_apply_func)
Out[7]:
A B
A
1 1 2
2 2 5
3 3 7
4 NaN NaN
For your reference the apply method automatically passes the data frame as the first argument.
Also, as you are always going to be reducing each group of data to a single observation you could also use the agg method (aggregate). apply is more flexible in terms of the length of the sequences that can be returned whereas agg must reduce the data to a single value.
df.groupby('A').agg(user_apply_func)
Related
I have a dataframe (small sample shown below, it has more columns), and I want to find the column names with the minimum values.
Right now, I have the following code to deal with it:
finaldf['min_pillar_score'] = finaldf.iloc[:, 2:9].idxmin(axis="columns")
This works fine, but does not return multiple values of column names in case there is more than one instance of minimum values. How can I change this to return multiple column names in case there is more than one instance of the minimum value?
Please note, I want row wise results, i.e. minimum column names for each row.
Thanks!
try the code below and see if it's in the output format you'd anticipated. it produces the intended result at least.
result will be stored in mins.
mins = df.idxmin(axis="columns")
for i, r in df.iterrows():
mins[i] = list(r[r == r[mins[i]]].index)
Get column name where value is something in pandas dataframe might be helpful also.
EDIT: adding an image of the output and the full code context.
Assuming this input as df:
A B C D
0 5 8 9 5
1 0 0 1 7
2 6 9 2 4
3 5 2 4 2
4 4 7 7 9
You can use the underlying numpy array to get the min value, then compare the values to the min and get the columns that have a match:
s = df.eq(df.to_numpy().min()).any()
list(s[s].index)
output: ['A', 'B']
Let's say I have a data like this:
df=pd.DataFrame({'a':[1,2,3,4,5,6,7,8,9,10,0,11,12],
'b':[0,0,0,0,0,0,0,0,0,0,0,0,1]})
I want to use b to put the data into two groups and sample from each. You can see that group 0 has much more data than group 1. So, if I do:
df1=df.groupby(['b']).apply(lambda x: x.sample(frac=0.1)).reset_index(drop=True)
You can find group 1 cannot be sampled. It might be sampled if frac increases.
So, what I should do to keep all the groups even it is very small?
Use sample reorder the dataframe , then we find the min count value per group , and the we can do head
df1 = df.groupby('b').apply(lambda x: x.sample(frac=1)).reset_index(drop=True)
ming = df1.b.value_counts().min()
df1 = df1.groupby('b').head(ming)
df1
Out[287]:
a b
0 8 0
12 12 1
I'm attempting to determine the number of widget failures from a test population.
Each widget can fail in 0, 1, or multiple ways. I'd like to calculate the number of failures of for each failure method, but once a widget is known to have failed, it should be excluded from future sums. In other words, the failure modes are known and ordered. If a widget fails via mode 1 and mode 3, I don't care about mode 3: I just want to count mode 1.
I have a dataframe with one row per item, and one column per failure mode. If the widget fails in that mode, the column value is 1, else it is 0.
d = {"item_1":
{"failure_1":0, "failure_2":0},
"item_2":
{"failure_1":1, "failure_2":0},
"item_3":
{"failure_1":0, "failure_2":1},
"item_4":
{"failure_1":1, "failure_2":1}}
df = pd.DataFrame(d).T
display(df)
Output:
failure_1 failure_2
item_1 0 0
item_2 1 0
item_3 0 1
item_4 1 1
If I just want to sum the columns, that's easy: df.sum(). And if I want to calculate percentage failures, easy too: df.sum()/len(df). But this counts widgets that fail in multiple ways, multiple times. For the problem stated, the best I can come up with is this:
# create empty df to store results
df2 = pd.DataFrame(columns=["total_failures"])
for col in df.columns:
# create a row, named after the column, and assign it the value of the sum
df2.loc[col] = df[col].sum()
# drop rows in the df column that are equal to 1
df = df.loc[df[col] != 1]
display(df2)
Output:
total_failures
failure_1 2
failure_2 1
This requires creating another dataframe (that's fine), but also requires iterating over the existing dataframe columns and deleting it a couple of rows at a time. If the dataframe takes a while to generate, or is needed for future calculations, this is not workable. I can deal with iterating over the columns.
Is there a way to do this without deleting the original df, or making a temporary copy? (Not workable with large data sets.)
You can do a cumsum on axis=1 and wherever the value is greater than 1 , mask it as 0 and then take sum:
out = df.mask(df.cumsum(axis=1).gt(1), 0).sum().to_frame('total_failures')
print(out)
total_failures
failure_1 2
failure_2 1
This way the original df is retained too.
I have a Pandas data frame which has some duplicate values, not rows. I want to use groupby.apply to remove the duplication. An example is as follows.
df = pd.DataFrame([['a', 1, 1], ['a', 1, 2], ['b', 1, 1]], columns=['A', 'B', 'C'])
A B C
0 a 1 1
1 a 1 2
2 b 1 1
# My function
def get_uniq_t(df):
if df.shape[0] > 1:
df['D'] = df.C * 10 + df.B
df = df[df.D == df.D.max()].drop(columns='D')
return df
df = df.groupby('A').apply(get_uniq_t)
Then I get the following value error message. The issue seems to do with creating the new column D. If I create the column D outside the function, the code seems running fine. Can someone help explain what caused the value error message?
ValueError: Shape of passed values is (3, 3), indices imply (2, 3)
The problem with your code is that it attempts to modify
the original group.
Other problem is that this function should return a single row
not a DataFrame.
Change your function to:
def get_uniq_t(df):
iMax = (df.C * 10 + df.B).idxmax()
return df.loc[iMax]
Then its application returns:
A B C
A
a a 1 2
b b 1 1
Edit following the comment
In my opinion, it is not allowed to modify the original group,
as it would indirectly modify the original DataFrame.
At least it displays a warning about this and is considered a bad practice.
Search the Web for SettingWithCopyWarning for more extensive description.
My code (get_uniq_t function) does not modify the original group.
It only returns one row from the current group.
The returned row is selected based on which row returns the greatest value
of df.C * 10 + df.B. So when you apply this function, the result is a new
DataFrame, with consecutive rows equal to results of this function
for consecutive groups.
You can perform an operation equivalent to modification, when you
create some new content, e.g. as the result of groupby instruction
and then save it under the same variable which so far held the source
DataFrame.
I have two dataframes as:
df1.ix[1:3]
DateTime
2018-01-02 [-0.0031537018416199097, 0.006451397621428631,...
2018-01-03 [-0.0028882814454597745, -0.005829869983964528...
df2.ix[1:3]
DateTime
2018-01-02 [-0.03285881500135208, -0.027806145786217932, ...
2018-01-03 [-0.0001314381449719178, -0.006278235444742629...
len(df1.ix['2018-01-02'][0])
500
len(df2.ix['2018-01-02'][0])
500
When I do df1 + df2 I get:
len((df1 + df2).ix['2018-01-02'][0])
1000
So, the lists instead of being summation is being concatenated.
How do I add element wise the lists in the dataframes df1 and df2.
When an operation is applied between two dataframes, it gets broadcasted at element level. Element in your case is a list and when '+' operator is applied between two lists, it concatenates them. That's why resulting dataframe contains concatenated lists.
There can be multiple approaches for actually summing up elements of lists instead of concatenating.
One approach can be converting list elements into columns and then adding dataframes and then merging columns back to a single list.(which has been suggested in first answer but in a wrong way)
Step 1: Converting list elements to columns
df1=df1.apply(lambda row:pd.Series(row[0]), axis=1)
df2=df2.apply(lambda row:pd.Series(row[0]), axis=1)
We need to pass row[0] instead of row to get rid of column index associated with series.
Step 2: Add dataframes
df=df1+df2 #this dataframe will have 500 columns
Step 3: Merge columns back to lists
df=df.apply(lambda row:pd.Series({0:list(row)}),axis=1)
This is an interesting part. Why are we returning a series here? Why only returning list(row) doesn't work and keep retaining 500 columns?
Reason is - if length of list returned is same as length of columns in the beginning, then this list gets fit in columns and to us it seems nothing happened. Whereas if length of the list is not equal to number of columns, then it is returned as single list.
Let's look at an example.
Suppose I've a dataframe, having columns 0 ,1 and 2.
df=pd.DataFrame({0:[1,2,3],1:[4,5,6],2:[7,8,9]})
0 1 2
0 1 4 7
1 2 5 8
2 3 6 9
Number of columns in original dataframe are 3. If I try to return a list with two columns, it works and a series is returned,
df1=df.apply(lambda row:[row[0],row[1]],axis=1)
0 [1, 4]
1 [2, 5]
2 [3, 6]
dtype: object
Instead if try to return list of three numbers, it would get fit in columns.
df1=df.apply(list,axis=1)
0 1 2
0 1 4 7
1 2 5 8
2 3 6 9
So if we want to return list of same size as number of columns, we'll have to return it in form of Series where one row's value has been given as list.
Another approach can be, introduce one column of a dataframe into other and then add columns using apply function.
df1[1]=df2[0]
df=df1.apply(lambda r: list(np.array(r[0])+np.array(r[1])),axis=1)
We can take advantage of numpy arrays here. '+' operator on numpy arrays sums up corresponding values and gives a single numpy array.
Cast them to series so that they become columns, then add your dfs:
df1 = df1.apply(pd.Series, axis=1)
df2 = df2.apply(pd.Series, axis=1)
df1 + df2