1 for循环语句如何同时获取值和索引

Question

I have a data frame and I would want to use for loop to get the column values and the index of that value.

Below is the dataframe and I am trying to get values from Date column

Below is my code. I declared a count variable to track the index. My question: Is it possible wherein the for loop declaration, I can get the column value and its index in one line?

Meaning in this line for row in loadexpense_df["Date"]: , row is the variable containing the value in the date column. Can improve the for loop to get value and its index?

Thanks

count =0
load = loadexpense_df["Date"]
for row in loadexpense_df["Date"]:
    checkMonth = row.strftime("%m")

    if checkMonth == '01':
        loadexpense_df["Month"][count] = "Jul"
    elif checkMonth == '02':
        loadexpense_df["Month"][count] = "Aug"
    elif checkMonth == '03':
        loadexpense_df["Month"][count] = "Sep"
    elif checkMonth == '04':

    count = count +1

Answer 1

Here's an example that could help you.

mylist = ["ball","cat","apple"]

for idx, val in enumerate(mylist):
    print ("index is {0} and value is {1}".format(idx, val))

Answer 2

iterrows returns the index and the row with the row represented as a series

for index, row in df.iterrows():

See here for more info:

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.iterrows.html

Answer 3

This is what iteritems is for:

for idx, val in loadexpense_df['Date'].items():
    pass

However, your code might have some problem with chain indexing. For example:

loadexpense_df["Month"][count] = "Jul"

I think you should look at np.select , or .loc access. That helps with code readability as well as performance. For example:

checkMonth = loadexpense_df['Date'].dt.month

loadexpense_df.loc[checkMonth==1, 'month'] = 'Jul'
...

Answer 4

You must think "pandas way", and the loop creation should be left to the pandas whenever possible. A solution example:

df
Date  Amount
0 2019-10-25       2
1 2019-10-26       5
2 2019-10-27      52
3 2019-10-28      93
4 2019-10-29      70
5 2019-10-30      51
6 2019-10-31      80
7 2019-11-01      61
8 2019-11-02      52
9 2019-11-03      61

m={10:'jul',11:'aug'}

# The easy way to get the month: dt.month
#df["Month"]= pd.Series.replace(df.Date.dt.month, m)
df["Month"]= df.Date.dt.month.replace(m)

        Date  Amount Month
0 2019-10-25       2   jul
1 2019-10-26       5   jul
2 2019-10-27      52   jul
3 2019-10-28      93   jul
4 2019-10-29      70   jul
5 2019-10-30      51   jul
6 2019-10-31      80   jul
7 2019-11-01      61   aug
8 2019-11-02      52   aug
9 2019-11-03      61   aug