How to delete rows with duplicate values in succeeding rows - pandas

in my csv data I have a column with a following data:
110.00
111.00
111.00 *
112.00
113.00
114.00
114.00 *
115.00
115.00 *
116.00
110.00
111.00
111.00 *
112.00
113.00
114.00
114.00 *
115.00
115.00 *
116.00
I read it in data frame and I'd like to delete on of the rows with the duplicating numbers but only if they are immedietely one after another. I marked the rows I's like to remove with an *.
Thanks for any suggestions

I think you can do this using .shift(), which can shift a series forward or backward (defaulting to one forward.) You want to keep rows if they're not the same as the next ones, so something like:
df[df["A"] != df["A"].shift()]
For example:
>>> df = pd.DataFrame({"A": [1,2,1,2,2,3,3,3,1,2]})
>>> df["A"]
0 1
1 2
2 1
3 2
4 2
5 3
6 3
7 3
8 1
9 2
Name: A, dtype: int64
>>> df["A"].shift()
0 NaN
1 1
2 2
3 1
4 2
5 2
6 3
7 3
8 3
9 1
Name: A, dtype: float64
>>> df["A"] != df["A"].shift()
0 True
1 True
2 True
3 True
4 False
5 True
6 False
7 False
8 True
9 True
Name: A, dtype: bool
Leading up to:
>>> df[df["A"] != df["A"].shift()]
A
0 1
1 2
2 1
3 2
5 3
8 1
9 2

Related

Assign column values from another dataframe with repeating key values

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')

Maximum of calculated pandas column and 0

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

Meaning of mode() in pandas

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

which rows are duplicates to each other

I have got a database with a lot of columns. Some of the rows are duplicates (on a certain subset).
Now I want to find out which row duplicates which row and put them together.
For instance, let's suppose that the data frame is
id A B C
0 0 1 2 0
1 1 2 3 4
2 2 1 4 8
3 3 1 2 3
4 4 2 3 5
5 5 5 6 2
and subset is
['A','B']
I expect something like this:
id A B C
0 0 1 2 0
1 3 1 2 3
2 1 2 3 4
3 4 2 3 5
4 2 1 4 8
5 5 5 6 2
Is there any function that can help me do this?
Thanks :)
Use DataFrame.duplicated with keep=False for mask with all dupes, then flter by boolean indexing, sorting by DataFrame.sort_values and join together by concat:
L = ['A','B']
m = df.duplicated(L, keep=False)
df = pd.concat([df[m].sort_values(L), df[~m]], ignore_index=True)
print (df)
id A B C
0 0 1 2 0
1 3 1 2 3
2 1 2 3 4
3 4 2 3 5
4 2 1 4 8
5 5 5 6 2

Pandas count values inside dataframe

I have a dataframe that looks like this:
A B C
1 1 8 3
2 5 4 3
3 5 8 1
and I want to count the values so to make df like this:
total
1 2
3 2
4 1
5 2
8 2
is it possible with pandas?
With np.unique -
In [332]: df
Out[332]:
A B C
1 1 8 3
2 5 4 3
3 5 8 1
In [333]: ids, c = np.unique(df.values.ravel(), return_counts=1)
In [334]: pd.DataFrame({'total':c}, index=ids)
Out[334]:
total
1 2
3 2
4 1
5 2
8 2
With pandas-series -
In [357]: pd.Series(np.ravel(df)).value_counts().sort_index()
Out[357]:
1 2
3 2
4 1
5 2
8 2
dtype: int64
You can also use stack() and groupby()
df = pd.DataFrame({'A':[1,8,3],'B':[5,4,3],'C':[5,8,1]})
print(df)
A B C
0 1 5 5
1 8 4 8
2 3 3 1
df1 = df.stack().reset_index(1)
df1.groupby(0).count()
level_1
0
1 2
3 2
4 1
5 2
8 2
Other alternative may be to use stack, followed by value_counts then, result changed to frame and finally sorting the index:
count_df = df.stack().value_counts().to_frame('total').sort_index()
count_df
Result:
total
1 2
3 2
4 1
5 2
8 2
using np.unique(, return_counts=True) and np.column_stack():
pd.DataFrame(np.column_stack(np.unique(df, return_counts=True)))
returns:
0 1
0 1 2
1 3 2
2 4 1
3 5 2
4 8 2