If I have a database
Example:
Name A B C
0 Jon 0 1 0
1 Jon 1 0 1
2 Alan 1 0 0
3 Shaya 0 1 1
If there is a duplicate in my dataset I want the person who has column A as 1 to have precedence.
NB. Column A can only have values 1 or 0
Output:
Name A B C
1 Jon 1 0 1
2 Alan 1 0 0
3 Shaya 0 1 1
IIUC sort value before drop duplicate
df.sort_values('A').drop_duplicates('Name',keep='last').sort_index()
Out[126]:
Name A B C
1 Jon 1 0 1
2 Alan 1 0 0
3 Shaya 0 1 1
Related
need some help here.. I am looking to retrieve Gender from Sheet 2 corresponding to the name in Sheet 1.
Step 1 - Match the name in sheet 1 to sheet 2 (not all names in sheet 1 will be in sheet 2, mark NA for non matching names)
Step 2 - Look for the corresponding gender in sheet 2.
Step 3 - Retrieve the column header or the last number in the column header (1,2,3...6)
Sheet 1
Name
Gender
w
???
e
r
t
y
u
i
q
w
e
r
Sheet 2
Name
Male 1
Female 2
other 3
other 4
other 5
Do not know 6
w
1
0
0
0
0
0
a
0
0
0
0
0
1
q
1
0
0
0
0
0
r
0
1
0
0
0
0
e
1
0
0
0
0
0
t
0
0
0
0
1
0
y
0
0
0
0
0
1
u
0
1
0
0
0
0
with Office 365 we can use FILTER:
=IFERROR(FILTER($F$1:$K$1,INDEX($F$2:$K$9,MATCH(A2,$E$2:$E$9,0),0)=1),"No Match")
With older versions we can use another INDEX/MATCH:
=IFERROR(INDEX($F$1:$K$1,MATCH(1,INDEX($F$2:$K$9,MATCH(A2,$E$2:$E$9,0),0),0)),"No Match")
To count the particular value of given column
Use pd.crosstab with df.sum:
In [236]: output = pd.crosstab(df['Rel_ID'], df['Values'])
In [238]: output['total'] = output.sum(axis=1)
In [239]: output
Out[239]:
Values 400.0 500.0 1700.0 6300.0 total
Rel_ID
TESTA 1 1 1 1 4
TESTB 1 0 1 1 3
TESTC 0 1 1 0 2
TESTD 1 0 1 1 3
TESTE 1 1 0 0 2
I have dataframe that has 50 columns each column have either 0 or 1. How do I return all rows that have an equal (tie) in the number of 0 and 1 (25 "0" and 25 "1").
An example on a 4 columns:
A B C D
1 1 0 0
1 1 1 0
1 0 1 0
0 0 0 0
based on the above example it should return the first and the third row.
A B C D
1 1 0 0
1 0 1 0
Because you have four columns, we assume you must have atleast two sets of 1 in a row. So, please try
df[df.mean(1).eq(0.5)]
I have a dataframe in pandas, an example of which is provided below:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6
A 1 0 0 1 0 0
B 1 1 0 0 1 0
C 1 0 1 1 0 0
D 0 0 1 0 0 1
E 1 1 1 1 1 1
As you can see 1 and 0 occurs randomly in different columns. It would be helpful, if anyone can suggest me a code in python such that I am able to find the column number where the 1 0 0 pattern occurs for the first time. For example, for member A, the first 1 0 0 pattern occurs at appear_1. so the first occurrence will be 1. Similarly for the member B, the first 1 0 0 pattern occurs at appear_2, so the first occurrence will be at column 2. The resulting table should have a new column named 'first_occurrence'. If there is no such 1 0 0 pattern occurs (like in row E) then the value in first occurrence column will the sum of number of 1 in that row. The resulting table should look something like this:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6 first_occurrence
A 1 0 0 1 0 0 1
B 1 1 0 0 1 0 2
C 1 0 1 1 0 0 4
D 0 0 1 0 0 1 3
E 1 1 1 1 1 1 6
Thank you in advance.
I try not to reinvent the wheel, so I develop on my answer to previous question. From that answer, you need to use additional idxmax, np.where, and get_indexer
cols = ['appear_1', 'appear_2', 'appear_3', 'appear_4', 'appear_5', 'appear_6']
df1 = df[cols]
m = df1[df1.eq(1)].ffill(1).notna()
df2 = df1[m].bfill(1).eq(0)
m2 = df2 & df2.shift(-1, axis=1, fill_value=True)
df['first_occurrence'] = np.where(m2.any(1), df1.columns.get_indexer(m2.idxmax(1)),
df1.shape[1])
Out[540]:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6 first_occurrence
0 A 1 0 0 1 0 0 1
1 B 1 1 0 0 1 0 2
2 C 1 0 1 1 0 0 4
3 D 0 0 1 0 0 1 3
4 E 1 1 1 1 1 1 6
I have following dataframe
A | B | C | D
1 0 2 1
0 1 1 0
0 0 0 1
I want to add the new column have any value of row in the column greater than zero along with column name
A | B | C | D | New
1 0 2 1 A-1, C-2, D-1
0 1 1 0 B-1, C-1
0 0 0 1 D-1
We can use mask and stack
s=df.mask(df==0).stack().\
astype(int).astype(str).\
reset_index(level=1).apply('-'.join,1).add(',').sum(level=0).str[:-1]
df['New']=s
df
Out[170]:
A B C D New
0 1 0 2 1 A-1,C-2,D-1
1 0 1 1 0 B-1,C-1
2 0 0 0 1 D-1
Combine the column names with the df values that are not zero and then filter out the None values.
df = pd.read_clipboard()
arrays = np.where(df!=0, df.columns.values + '-' + df.values.astype('str'), None)
new = []
for array in arrays:
new.append(list(filter(None, array)))
df['New'] = new
df
Out[1]:
A B C D New
0 1 0 2 1 [A-1, C-2, D-1]
1 0 1 1 0 [B-1, C-1]
2 0 0 0 1 [D-1]