df5 = pd.DataFrame({"A": np.random.randint(0, 7, size=50),
"B": np.random.randint(-10, 15, size=50)})
df5.mode()
A B
0 1.0 -9
1 NaN 10
2 NaN 13
Why does the NaN come from here?
Reason is if check DataFrame.mode:
Get the mode(s) of each element along the selected axis.
The mode of a set of values is the value that appears most often. It can be multiple values.
So missing values means for A is ony one mode value, for B column are 3 mode values, so for same rows are added missing values.
If check my sample data - there is mode A 2 times and B only once, because 2and 3 are both 11 times in data:
np.random.seed(20)
df5 = pd.DataFrame({"A": np.random.randint(0, 7, size=50),
"B": np.random.randint(-10, 15, size=50)})
print (df5.mode())
A B
0 2 8.0
1 3 NaN
print (df5.A.value_counts())
3 11 <- both top1
2 11 <- both top1
6 9
5 8
0 5
1 4
4 2
Name: A, dtype: int64
print (df5.B.value_counts())
8 6 <- only one top1
0 4
4 4
-4 3
10 3
-2 3
1 3
12 3
6 3
7 2
3 2
5 2
-9 2
-6 2
14 2
9 2
-1 1
11 1
-3 1
-7 1
Name: B, dtype: int64
Related
I have a data frame with three column variables A,B,C, taking numeric values in {1,2}, {6,7}, and {11,12}. I would like to see the following. For what fraction of possible observed pairs (A,B) do we have both [observations for which C=11 and observations for which C=12].
I start by entering the dataframe:
df = pd.DataFrame({"A": [1, 2, 1, 1, 2, 1, 1, 2], "B": [6,7,7,6,7,6,6,6], "C": [11,12,11,11,12,12,11,12]})
--------
A B C
0 1 6 11
1 2 7 12
2 1 7 11
3 1 6 11
4 2 7 12
5 1 6 12
6 1 6 11
7 2 6 12
Then I think I need to use groupby. I run
g = df.groupby(["A", "B"])
"g.C.value_counts()"
-----------
A B C
1 6 11 3
12 1
7 11 1
2 6 12 1
7 12 2
Name: C, dtype: int64
This shows that we have one pair of (A,B) for which we have both a C=11 and a C=12, and 3 pairs of (A,B) for which we only have either C=11 or C=12. So I would like to make pandas tells me that we have 25% of (A,B) paris for which C takes both values and 75% for which it only takes one value.
How can I accomplish this? I would like to do so for a big data frame where I can't just eyeball it from the value_counts--this small dataframe is just to illustrate.
Thanks!
Pass normalize=True
out = df.groupby(["A", "B"]).C.value_counts(normalize=True)
Out[791]:
A B C
1 6 11 0.75
12 0.25
7 11 1.00
2 6 12 1.00
7 12 1.00
Name: C, dtype: float64
Please help me in Pandas, i cant find good solution
Tried map, assign, merge, join, set_index.
Maybe just i am too tired :)
df:
m_num A B
0 1 0 9
1 1 1 8
2 2 2 7
3 2 3 6
4 3 4 5
5 3 5 4
df1:
m_num C
0 2 99
1 2 88
df_final:
m_num A B C
0 1 0 9 NaN
1 1 1 8 NaN
2 2 2 7 99
3 2 3 6 88
4 3 4 5 NaN
5 3 5 4 NaN
Try:
df2 = df[df['m_num'].isin(df1['m_num'])].reset_index(drop=True)
df2 = pd.merge(df2,df1,on=[df1.index,'m_num']).drop('key_0',axis=1)
df2 = pd.merge(df,df2,on=['m_num','A','B'],how='left')
print(df2)
Prints:
m_num A B C
0 1 0 9 NaN
1 1 1 8 NaN
2 2 2 7 99.0
3 2 3 6 88.0
4 3 4 5 NaN
5 3 5 4 NaN
Explanation:
There may be better solutions out there but this was my thought process. The problem is slightly tricky in the sense that because 'm_num' is the only common key and it and it has repeating values.
So first I created a dataframe matching df and df1 here so that I can use the index as another key for the subsequent merge.
df2 = df[df['m_num'].isin(df1['m_num'])].reset_index(drop=True)
This prints:
m_num A B
0 2 2 7
1 2 3 6
As you can see above, now we have the index 0 and 1 in addition to the m_num as key which we can use to match with df1.
df2 = pd.merge(df2,df1,on=[df1.index,'m_num']).drop('key_0',axis=1)
This prints:
m_num A B C
0 2 2 7 99
1 2 3 6 88
Then tie the above resultant dataframe to the original df and do a left join to get the output.
df2 = pd.merge(df,df2,on=['m_num','A','B'],how='left')
I have a very simple problem (I guess) but don't find the right syntax to do it :
The following Dataframe :
A B C
0 7 12 2
1 5 4 4
2 4 8 2
3 9 2 3
I need to create a new column D equal for each row to max (0 ; A-B+C)
I tried a np.maximum(df.A-df.B+df.C,0) but it doesn't match and give me the maximum value of the calculated column for each row (= 10 in the example).
Finally, I would like to obtain the DF below :
A B C D
0 7 12 2 0
1 5 4 4 5
2 4 8 2 0
3 9 2 3 10
Any help appreciated
Thanks
Let us try
df['D'] = df.eval('A-B+C').clip(lower=0)
Out[256]:
0 0
1 5
2 0
3 10
dtype: int64
You can use np.where:
s = df["A"]-df["B"]+df["C"]
df["D"] = np.where(s>0, s, 0) #or s.where(s>0, 0)
print (df)
A B C D
0 7 12 2 0
1 5 4 4 5
2 4 8 2 0
3 9 2 3 10
To do this in one line you can use apply to apply the maximum function to each row seperately.
In [19]: df['D'] = df.apply(lambda s: max(s['A'] - s['B'] + s['C'], 0), axis=1)
In [20]: df
Out[20]:
A B C D
0 0 0 0 0
1 5 4 4 5
2 0 0 0 0
3 9 2 3 10
I have a dataframe:
np.random.seed(1)
df1 = pd.DataFrame({'day':[3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6],
'item': [1, 1, 2, 2, 1, 2, 3, 3, 4, 3, 4],
'price':np.random.randint(1,30,11)})
day item price
0 3 1 6
1 4 1 12
2 4 2 13
3 4 2 9
4 5 1 10
5 5 2 12
6 5 3 6
7 5 3 16
8 5 4 1
9 6 3 17
10 6 4 2
After the groupby code gb = df1.groupby(['day','item'])['price'].mean(), I get:
gb
day item
3 1 6
4 1 12
2 11
5 1 10
2 12
3 11
4 1
6 3 17
4 2
Name: price, dtype: int64
I want to get the trend from the groupby series replacing back into the dataframe column price. The price is the variation of the item-price with repect to the previous day price
day item price
0 3 1 nan
1 4 1 6
2 4 2 nan
3 4 2 nan
4 5 1 -2
5 5 2 1
6 5 3 nan
7 5 3 nan
8 5 4 nan
9 6 3 6
10 6 4 1
Please help me to code the last step. A single/double line code will be most helpful. As the actual dataframe is huge, I would like to avoid iterations.
Hope this helps!
#get the average values
mean_df=df1.groupby(['day','item'])['price'].mean().reset_index()
#rename columns
mean_df.columns=['day','item','average_price']
#sort by day an item in ascending
mean_df=mean_df.sort_values(by=['day','item'])
#shift the price for each item and each day
mean_df['shifted_average_price'] = mean_df.groupby(['item'])['average_price'].shift(1)
#combine with original df
df1=pd.merge(df1,mean_df,on=['day','item'])
#replace the price by difference of previous day's
df1['price']=df1['price']-df1['shifted_average_price']
#drop unwanted columns
df1.drop(['average_price', 'shifted_average_price'], axis=1, inplace=True)
I have a dataframe in the shape of [100, 50000]
and I want to reduce it by applying mean per row in chunks of 5. (So I will get a dataframe at the shape of [100, 10000]).
For example,
So, if the row is
[1,8,-1,0,2 , 6,8,11,4,6]
the output will be
[2,7]
What is the most efficient way to do so?
Thanks
If shape is 100, 50000 means 100 rows and 50000 columns, solution is GroupBy.mean with helper np.arange created by lengths of columns and axis=1:
df = pd.DataFrame([[1,8,-1,0,2 , 6,8,11,4,6],
[1,8,-1,0,2 , 6,8,11,4,6]])
print (df)
0 1 2 3 4 5 6 7 8 9
0 1 8 -1 0 2 6 8 11 4 6
1 1 8 -1 0 2 6 8 11 4 6
print (df.shape)
(2, 10)
df = df.groupby(np.arange(len(df.columns)) // 5, axis=1).mean()
print (df)
0 1
0 2 7
1 2 7
If shape is 100, 50000 means 100 columns and 50000 rows, solution is GroupBy.mean with helper np.arange created by lengths of DataFrame:
df = pd.DataFrame({'a': [1,8,-1,0,2 , 6,8,11,4,6],
'b': [1,8,-1,0,2 , 6,8,11,4,6]})
print (df)
a b
0 1 1
1 8 8
2 -1 -1
3 0 0
4 2 2
5 6 6
6 8 8
7 11 11
8 4 4
9 6 6
print (df.shape)
(10, 2)
df = df.groupby(np.arange(len(df)) // 5).mean()
print (df)
a b
0 2 2
1 7 7