Python pandas 如果行包含字符串添加到表

Question

I have a spreadsheet that looks as follows:

my goal is to print out which "hmi" controls which "az", which is indicated by an "x" in its corresponding column. Example based on the above spreadsheet:

• HMI1 controls AZ1
• HMI2 controls AZ2
• HMI3 controls AZ3
• HMI4 controls AZ3

etc..

My idea output that i can work with and sort to an output excel sheet would be(where "X" is the corresponding number):

HMIX | AZX

ive imported the spreadsheet with pd.read_excel and my thought was using the following code to format it, however i think im close but on the wrong path:

    for index, row in dfConfig.iterrows():
        if row["az1"] != "NaN":
            print(row["hmi_number"])

Answer 1

Given df,

df = pd.DataFrame({'hmi_number':np.arange(1,10),
                  'az1':['x']+[np.nan]*8,
                  'az2':[np.nan]+['x']+[np.nan]*7,
                  'az3':[np.nan]*2+['x']*2+[np.nan]*5,
                  'az4':[np.nan]*4+['x']*2+[np.nan]*3,
                  'az5':[np.nan]*6+['x']*2+[np.nan],
                  'az6':[np.nan]*8+['x']})

Try this:

for hmi, az in (df.set_index('hmi_number')=='x').dot(df.columns[1:]).iteritems():
    print(f'HMI{hmi} controls {az}')

Output:

HMI1 controls az1
HMI2 controls az2
HMI3 controls az3
HMI4 controls az3
HMI5 controls az4
HMI6 controls az4
HMI7 controls az5
HMI8 controls az5
HMI9 controls az6

Answer 2

Iterate over rows first, thank over columns, if you find the x , print the row and the columns.

for i, row in dfConfig.iterrows():
    for col in dfConfig:
        if row[col] == 'x':
            print(f'HMI{row["hmi_number"]} controls {col}')

Answer 3

If your looking for something that checks if something is a string try this

>>> currentColumn = "x"
>>> isinstance(currentColumn, str)
True
>>> isinstance(currentColumn, int)
False
>>> 

Answer 4

Another approach is based on stack() method and then accessing index by list comprehension. Given above dataframe df :

[print(f"HMI{x} controls {y}") for x,y in df.set_index("hmi_number").stack().index]

Solution is fast and very compact.