Here are the tables before cleaned:
name
date
time_lag1
time_lag2
time_lag3
lags
a
2000/5/3
1
0
1
time_lag1
a
2000/5/10
1
1
0
time_lag2
a
2000/5/17
1
1
1
time_lag3
b
2000/5/3
0
1
0
time_lag1
c
2000/5/3
0
0
0
time_lag1
Logics are simple, each name have several date and that date correspond to a "lags". What I tried to do is to match the column names like "time_lag1","time_lag2",...,"time_lagn" to the values in column "lags". For example, the first value of "time_lag1" is because column name "time_lag1" equals the corresponding value of "lags" which is also "time_lag1". However, I don't know why the values of other columns and rows are becoming incorrect.
My thought is:
# time_lag columns are not following a trend, so it can be lag_time4 as well.
time_list = ['time_lag1','time_lag2','lag_time4'...]
for col in time_list:
if col == df['lags'].values:
df.col == 1
else:
df.col == 0
I don't know why the codes I tried is not working very well.
Here are the tables I tried to get:
name
date
time_lag1
time_lag2
time_lag3
lags
a
2000/5/3
1
0
0
time_lag1
a
2000/5/10
0
1
0
time_lag2
a
2000/5/17
0
0
1
time_lag3
b
2000/5/3
1
0
0
time_lag1
c
2000/5/3
1
0
0
time_lag1
The simplest is to recalculate them from scratch with pandas.get_dummies and to update the dataframe:
df.update(pd.get_dummies(df['lags']))
Output:
name date time_lag1 time_lag2 time_lag3 lags
0 a 2000/5/3 1 0 0 time_lag1
1 a 2000/5/10 0 1 0 time_lag2
2 a 2000/5/17 0 0 1 time_lag3
3 b 2000/5/3 1 0 0 time_lag1
4 c 2000/5/3 1 0 0 time_lag1
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")
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 time series dataframe with the below columns. I am trying to figure out:
If df['PH'] ==1, then I need find the previous date where df['pivot_low_1'] == 1 and extract the value of df['low'] for that date. So, for 2010-01-12 where df['PH'] ==1, I would need to identify the previous non-zero df['pivot_low_1'] == 1 on 2010-01-07 and get df['low'] == 1127.00000.
low pivot_low_1 PH
date
2010-01-04 1114.00000 1 0
2010-01-05 1125.00000 0 0
2010-01-06 1127.25000 0 0
2010-01-07 1127.00000 1 0
2010-01-08 1131.00000 0 0
2010-01-11 1137.75000 0 0
2010-01-12 1127.75000 1 1
2010-01-13 1129.25000 0 0
2010-01-14 1138.25000 0 0
2010-01-15 1127.50000 1 0
2010-01-18 1129.50000 0 0
2010-01-19 1126.25000 0 0
2010-01-20 1125.25000 0 0
2010-01-21 1108.50000 0 0
2010-01-22 1086.25000 1 0
2010-01-25 1089.75000 0 0
2010-01-26 1081.00000 0 0
2010-01-27 1078.50000 0 0
2010-01-28 1074.25000 0 0
2010-01-29 1066.50000 1 1
2010-02-01 1068.00000 0 0
since you want a column in same dataframe but the output is correspondent to only certain rows , I will be replacing every other column with nan values,
data = pd.read_csv('file.csv')
data.columns=['low', 'pivot_low_1', 'PH']
count = 0
l = list()
new=list()
for index, row in data.iterrows():
if row['pivot_low_1']==1:
l.append(count)
if (row['PH']==1) and (row['pivot_low_1']==1):
new.append(data.iloc[l[len(l)-2]].low)
elif (row['PH']==1):
new.append(data.iloc[l[len(l)-1]].low)
elif (row['PH']==0):
new.append(np.nan)
count+=1
data['new'] = new
data
The output is as shown in this image, https://imgur.com/a/IqowZHZ , hope this helps
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 a dataset 'df' that looks something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
A 1 0 0 1 0 1
B 1 1 0 0 1 0
C 1 1 1 0 0 1
D 0 0 1 0 0 1
As you can see there are several rows of ones and zeros. Can anyone suggest me a code in python such that I am able to count the number of times '1' occurs continuously before the first occurrence of a 1, 0 and 0 in order. For example, for member A, the first double zero event occurs at seen_2 and seen_3, so the event will be 1. Similarly for the member B, the first double zero event occurs at seen_3 and seen_4 so there are two 1s that occur before this. The resultant table should have a new column 'event' something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
A 1 0 0 1 0 1 1
B 1 1 0 0 1 0 2
C 1 1 1 0 0 1 3
D 0 0 1 0 0 1 1
My approach:
df = df.set_index('MEMBER')
# count 1 on each rows since the last 0
s = (df.stack()
.groupby(['MEMBER', df.eq(0).cumsum(1).stack()])
.cumsum().unstack()
)
# mask of the zeros:
u = s.eq(0)
# look for the first 1 0 0
idx = (~u &
u.shift(-1, axis=1, fill_value=False) &
u.shift(-2, axis=1, fill_value=False) ).idxmax(1)
# look up
df['event'] = s.lookup(idx.index, idx)
Test data:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
0 A 1 0 1 0 0 1
1 B 1 1 0 0 1 0
2 C 1 1 1 0 0 1
3 D 0 0 1 0 0 1
4 E 1 0 1 1 0 0
Output:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
0 A 1 0 1 0 0 1 1
1 B 1 1 0 0 1 0 2
2 C 1 1 1 0 0 1 3
3 D 0 0 1 0 0 1 1
4 E 1 0 1 1 0 0 2