Question

我有一个看起来像这样的DataFrame：

columns = ['C24H31O8', 'C23H27O9', 'C22H23O10', 'C21H19O11', 'C20H15O12','C19H11O13']

values= [[   0.,   37.,   74.,  111.,  148.,  185.],
         [ -37.,    0.,   37.,   74.,  111.,  148.],
         [ -74.,  -37.,    0.,   37.,   74.,  111.],
         [-111.,  -74.,  -37.,    0.,   37.,   74.],
         [-148., -111.,  -74.,  -37.,    0.,   37.],
         [-185., -148., -111.,  -74.,  -37.,    0.]]

您可以使用轻松地重新创建我的数据框

df = pd.DataFrame(data=values, index=columns)

我要遍历数据框，如下所示：

进行迭代时，在C24H31O8行中的第1列中找到了37。现在，转到第1行，然后再次进行迭代。如果您再次找到数字37（在这种情况下，您在第三列中找到了37），请转到第三行，然后再次搜索37，依此类推。

我想要的输出应该是一条链：

37 : C24H31O9 --> C23H27O9 --> C22H23O10 --> C21H19O11 ...

（为每个值做些事情）

Answer 1

如果我正确理解了这个问题，这应该可以工作：

columns= pd.Index(['C24H31O8', 'C23H27O9', 'C22H23O10', 'C21H19O11', 'C20H15O12','C19H11O13'],dtype='object')

values=np.array([[   0.,   37.,   74.,  111.,  148.,  185.],
               [ -37.,    0.,   37.,   74.,  111.,  148.],
               [ -74.,  -37.,    0.,   37.,   74.,  111.],
               [-111.,  -74.,  -37.,    0.,   37.,   74.],
               [-148., -111.,  -74.,  -37.,    0.,   37.],
               [-185., -148., -111.,  -74.,  -37.,    0.]])

df = pd.DataFrame(values, columns)

def build_string(df, i):
    row = 0
    chain = df.index[0]
    while True:
        try:
            row = df.iloc[row][df.iloc[row] == i].index[0]
            chain += f"--> {df.index[row]}"
        except IndexError:
            break
    return chain

输出：

>>> build_string(df, 37)
'C24H31O8--> C23H27O9--> C22H23O10--> C21H19O11--> C20H15O12--> C19H11O13'

或

>>> {int(i): build_string(df, i) for i in df.iloc[0] if i > 0}
{37: 'C24H31O8--> C23H27O9--> C22H23O10--> C21H19O11--> C20H15O12--> C19H11O13',
 74: 'C24H31O8--> C22H23O10--> C20H15O12',
 111: 'C24H31O8--> C21H19O11',
 148: 'C24H31O8--> C20H15O12',
 185: 'C24H31O8--> C19H11O13'}

Answer 2

IIUC：

#!/usr/bin/env bash
#--------find apk---------
apkFile=$(find -name '*.apk' -print | grep -oP '(?<=.).*(.apk)')

#--------find apkFilePath---------
if test -z "apkFile"
then
echo "apkFile: is NULL"
exit 0;
fi

echo "apkFile: ${apkFile}"
apkFilePath=$(pwd)${apkFile}
echo "apk file path: ${apkFilePath}"

#--------install---------
if test -z "$apkFilePath"
then
echo "apkFilePath: is NULL"
exit 0;
fi

echo "adb install -t -r ${apkFilePath}"
for SERIAL in $(adb devices | grep -v List | cut -f 1);
do `adb -s ${SERIAL} install -t -r ${apkFilePath}`;
done

#--------get applicationId---------
echo "aapt dump badging ${apkFilePath} | grep -oP '(?<=package: name=).*(?=versionCode)'"
applicationId=$(aapt dump badging ${apkFilePath} | grep -oP '(?<=package: name=).*(?=versionCode)')
echo "applicationId: is ${applicationId}"

#--------launch---------
if test -z "$applicationId"
then
echo "applicationId: is NULL"
exit 0;
fi

echo "____________________START_APPLICATION_ID________________________"
echo "applicationId: ${applicationId}"
echo "____________________END_APPLICATION_ID__________________________"
echo "____________________START_LAUNCHER______________________________"
for SERIAL in $(adb devices | grep -v List | cut -f 1);
do `adb -s ${SERIAL} shell monkey -p ${applicationId} -c android.intent.category.LAUNCHER 1`;
done
echo "____________________END_LAUNCHER________________________________"

输出（df = pd.DataFrame(values, columns = columns) s = df.where(df>0).stack() out_dict = s.index.get_level_values(1).groupby(s)）：

out_dict

遍历Pandas DataFrame并存储结果

2 个答案: